## Tag Archives: semidirect product

### Construct a nonsimple group of order 168 with more than one Sylow 7-subgroup

Prove or construct a counterexample to the assertion: if $G$ is a group of order 168 with more than one Sylow 7-subgroup, then $G$ is simple.

Recall that $GL_3(\mathbb{F}_2)$ has order $7 \cdot 6 \cdot 4$, and thus by Cauchy there exists a nontrivial group homomorphism $\varphi : Z_7 \rightarrow \mathsf{Aut}(Z_2^3)$. Then $Z_3 \times (Z_2^3 \rtimes_\varphi Z_7)$ has order 168, is not simple, and does not have a normal Sylow 7-subgroup by Proposition 5.11 and this previous exercise.

### Compute the isomorphism type of the Heisenberg group over ZZ/(p)

Let $H(\mathbb{F}_p)$ be the Heisenberg group over $\mathbb{F}_p = \mathbb{Z}/(p)$. Prove that $H(\mathbb{F}_2) \cong D_8$, and that $H(\mathbb{F}_p)$ has exponent $p$ and is isomorphic to the first nonabelian group in Example 7 in the text.

Define $A,B \in H(\mathbb{F}_2)$ by $A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Evidently $A^{-1} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ and $B^{-1} = B$. Moreover, $B \notin \langle A \rangle$, $A^4 = B^2 = 1$, and $AB = BA^{-1}$. By Lemma 1 to a previous exercise, we have an injective group homomorphism $\varphi : D_8 \rightarrow H(\mathbb{F}_2)$ such that $\varphi(r) = A$ and $\varphi(s) = B$. Since these groups are finite and have the same cardinality, $\varphi$ is an isomorphism.

Note that $H(\mathbb{F}_p)$ is a nonabelian group of order $p^3$; thus it is isomorphic to either $(Z_p \times Z_p) \rtimes Z_p$ or $Z_{p^2} \rtimes Z_p$. Now in the classification of groups of order $p^3$, we saw that $(Z_p \times Z_p) \rtimes Z_p$ has no elements of order $p^2$.

Now in a previous exercise we showed (over $\mathbb{R}$, but the proof holds over any field) that for all matrices $A = \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}$ in $H(\mathbb{F}_p)$, $A^p = \begin{bmatrix} 1 & pa & pb + p \frac{p+1}{2} ac \\ 0 & 1 & pc \\ 0 & 0 & 1 \end{bmatrix}$. Note that $(p+1)/2$ is an integer; thus $A^p = 1$. So $H(\mathbb{F}_p)$ has exponent $p$, and thus has no elements of order $p^2$. Hence $H(\mathbb{F}_p) \cong (Z_p \times Z_p) \rtimes Z_p$.

### The group of upper triangular matrices over a field is the semidirect product of the diagonal matrices by upper triangular matrices of determinant 1

Let $F$ be a field, let $n$ be a positive integer, and let $G = UT_n(F)$ be the group of upper triangular $n \times n$ matrices over $F$.

1. Prove that $G = U \rtimes D$, where $U = \overline{UT}_n(F)$ is the group of strictly upper triangular matrices and $D$ is the set of diagonal matrices in $GL_n(F)$.
2. Let $n = 2$. Recall that $U \cong F$ and $D \cong F^\times \times F^\times$. Describe the homomorphism $D \rightarrow \mathsf{Aut}(U)$ explicitly in terms of these isomorphisms. That is, show how each element of $F^\times \times F^\times$ acts as an automorphism of $F$.

1. Recall that $SL_n(F) \leq GL_n(F)$ is normal; thus $U = G \cap SL_n(F)$ is normal. Now let $M = [m_{i,j}] \in G$, and define $A \in U$ and $B \in D$ as follows: $A = [a_{i,j}]$, where $a_{i,j} = 1$ if $i=j$, $m_{i,j}/m_{j,j}$ if $ij$, then for all $latex k j$, $b_{k,j} = 0$. Thus $c_{i,j} = 0$. If $i = j$, then for all $latex k j$, $b_{k,j} = 0$. Thus $c_{i,j} = m_{i,j}$. If $i, then for all $k, $a_{i,k} = 0$, for $i \leq k j$, $b_{k,j} = 0$. Thus $c_{i,j} = a_{i,j}b_{j,j}$ $= (m_{i,j}/m_{j,j})m_{j,j} = m_{i,j}$. Thus we have $AB = M$. By the recognition theorem for semidirect products, $G = U \rtimes D$.
2. Let $B = \begin{bmatrix} \alpha & 0 \\ 0 & \beta \end{bmatrix} \in D$ and $A = \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix} \in U$. $D$ acts on $U$ by conjugation, and we have $BAB^{-1} = \begin{bmatrix} 1 & \alpha x \beta^{-1} \\ 0 & 1 \end{bmatrix}$. Thus the homomorphism $\varphi : F^\times \times F^\times \rightarrow F$ is given by $\varphi(\alpha,\beta)(x) = \alpha x \beta^{-1}$.

### Construct all the semidirect products of Cyc(2ⁿ) by Cyc(2)

Show that for any $n \geq 3$ there are exactly 4 distinct homomorphisms from $Z_2$ into $\mathsf{Aut}(Z_{2^n})$. Prove that the resulting semidirect products are nonisomorphic groups of order $2^{n+1}$. (These four groups, with the cyclic group and the generalized quaternion group, are all the groups of order $2^{n+1}$ which have a cyclic subgroup of index 2.)

We begin with two lemmas.

Lemma 1: Let $k \geq 3$ be an integer. Suppose $m$ and $n$ are integers such that, for all integers $a$ and $b$, $m+a(1+2^{k-1})^n \equiv a+m(1+2^{k-1})^b$ mod $2^k$. Then $m \equiv n \equiv 0$ mod 2. Proof: Note that $(1+2^{k-1})^2 \equiv 1$ mod $2^k$. Suppose $n \equiv 1$ mod 2. Let $b = 2$. Then we have $a2^{k-1} \equiv 0$ mod $2^k$, a contradiction when $a$ is odd. Thus $n \equiv 0$ mod $2^k$. Now choosing $b$ odd, we have $m2^{k-1} \equiv 0$ mod $2^k$; thus $m \equiv 0$ mod 2. $\square$

Lemma 2: Let $k \geq 3$ be an integer. Suppose $m$ and $n$ are integers such that, for all integers $a$ and $b$, $m+a(-1+2^{k-1})^n \equiv a+m(-1+2^{k-1})^b$ mod $2^k$. Then $m \equiv 0$ mod $2^{k-1}$ and $n \equiv 0$ mod 2. Proof: Note that $(-1+2^{k-1})^2 \equiv 1$ mod $2^k$. Suppose $n \equiv 1$ mod 2. Let $b = 2$. Then we have $0 \equiv 2a(1-2^{k-2})$ mod $2^k$. Since $1-2^{k-2}$ is odd, we have $2a \equiv 0$ mod $2^k$ for all $a$, which is absurd. Thus $n \equiv 0$ mod 2. Now let $b$ be odd; we have $2m(1-2^{k-2}) \equiv 0$ mod $2^k$; again since $1-2^{k-2}$ is odd we have $2m \equiv 0$ mod $2^k$, so that $m \equiv 0$ mod $2^{k-1}$. $\square$

We found that there are 4 distinct homomorphisms $Z_2 \rightarrow \mathsf{Aut}(Z_{2^n})$ in two lemmas to the previous exercise. Moreover, writing $Z_2 = \langle x \rangle$ and $Z_{2^n} = \langle y \rangle$, these are given as follows. $\varphi_1(x)(y) = y$, $\varphi_2(x)(y) = y^{-1}$, $\varphi_3(x)(y) = y^{1+2^{n-1}}$, and $\varphi_4(x)(y) = y^{-1+2^{n-1}}$.

Let $G_i = Z_{2^n} \rtimes_{\varphi_i} Z_2$. We will now show that these $G_i$ are distinct.

Note that $\varphi_1$ is trivial, so that $G_1 \cong Z_{2^n} \times Z_2$ is abelian. However, the remaining groups are not abelian, so $G_1$ is distinct from the remaining three.

Note that $G_2 \cong D_{2^{n+1}} = \langle r,s \rangle$. By this previous exercise, $Z(D_{2^{n+1}}) = \langle r^{2^{n-1}} \rangle \cong Z_2$.

Now $G_3$ is generated by $x$ and $y$, and we have $xyx^{-1} = \varphi_3(x)(y) = y^{1+2^{n-1}}$. Thus this group has the presentation $\langle x,y \ |\ x^2 = y^{2^n} = 1, xy = y^{1+2^{n-1}}x \rangle$. Similarly, $G_4$ has the presentation $\langle x,y \ |\ x^2 = y^{2^n} = 1, xy = y^{-1+2^{n-1}}x \rangle$.

We now compute the centers of $G_3$ and $G_4$.

Every element of $G_3$ can be written as $y^ix^j$ where $0 < i \leq 2^n$ and $0 < j \leq 2$, and there are $2^{n+1}$ such expressions. Since we know by other means that $|G_3| = 2^{n+1}$, every element of $G_3$ can be written in this form uniquely. Let $s = y^ax^b \in Z(G_3)$ and let $y^ix^j \in G_3$. Then $y^ax^by^ix^j = y^{a+i(1+2^{n-1})^b}x^{b+j}$ is equal to $y^ix^jy^ax^b = y^{i+a(1+2^{n-1})^j}x^{b+j}$. Comparing exponents and using Lemma 1, we have $a \equiv b \equiv 0$ mod 2. Thus $s \in \langle y^2 \rangle$. Now note that $y^2y^ix^j = y^{i+2}x^j$ and $y^ixy^2 = y^iy^{2(1+2^{n-1})}x = y^{i+2}x$, so that $y^2 \in Z(G_3)$; thus $Z(G_3) = \langle y^2 \rangle \cong Z_{2^{n-1}}$.

Performing the same analysis on $G_4$ reveals that $Z(G_4) = \langle y^{2^{n-1}} \rangle \cong Z_2$.

Thus $G_3$ is distinct from $G_2$ and $G_4$.

We know that $G_2 \cong D_{2^{n+1}}$ has $2^n + 1$ elements of order 2. Suppose $s = y^ax^b \in G_4$ has order 2; then $(y^ax^b)^2 = y^{a(1+(-1+2^{n-1})^b)} = 1$. If $b = 0$, then $a = 0$, so that $s = 1$. If $b = 1$, then $a2^{n-1} \equiv 0$ mod $2^n$, so that $a \equiv 0$ mod 2. There are $2^{n-1}$ such numbers between 0 and $2^n$. Thus $G_4$ has $2^{n-1}$ elements of order 2, and $G_2 \not\cong G_4$.

### Construct all the semidirect products of Cyc(2) by Cyc(8)

Prove that there are exactly 4 distinct homomorphisms $Z_2 \rightarrow \mathsf{Aut}(Z_8)$. Prove that the resulting semidirect products are $Z_8 \times Z_2$, $D_{16}$, $QD_{16}$, and $M$.

We begin with some lemmas.

Lemma 1: If $n \geq 3$, then $\mathsf{Aut}(Z_{2^n}) \cong Z_{2^{n-2}} \times Z_2$. Proof: We have $\mathsf{Aut}(Z_{2^n}) \cong \mathbb{Z}/(2^n)^\times$, and this group has order $2^{n-1}$. By this previous exercise, 5 has (multiplicative) order $2^{n-2}$ in $\mathbb{Z}/(2^n)^\times$. Using this previous exercise, -1 has order 2 in $\mathbb{Z}/(2^n)^\times$ and $-1 \notin \langle 5 \rangle$. Since $\mathbb{Z}/(2^n)^\times$ is abelian, both $\langle 5 \rangle$ and $\langle -1 \rangle$ are normal. Moreover, $\mathbb{Z}/(2^n)^\times = \langle 5 \rangle \langle -1 \rangle$; by the recognition theorem for direct products, $\mathsf{Aut}(Z_{2^n}) \cong \langle 5 \rangle \times \langle -1 \rangle \cong Z_{2^{n-2}} \times Z_2$. $\square$

Lemma 2: If $n \geq 3$, then there are exactly 4 distinct homomorphisms $\varphi : Z_2 \rightarrow \mathsf{Aut}(Z_{2^n})$. Proof: The distinct homomorphisms correspond precisely to elements of order 1 or 2 in $\mathsf{Aut}(Z_{2^n})$. By Lemma 1, $\mathsf{Aut}(Z_{2^n}) \cong \langle 5 \rangle \times \langle -1 \rangle = \mathbb{Z}/(2^n)^\times$. Since $|(x,y)| = \mathsf{lcm}(|x|,|y|)$, and $\langle 5 \rangle$ and $\langle -1 \rangle$ have the unique elements of order 2 $5^{2^{n-3}}$ and $-1$, respectively, the elements of order 2 in $\mathsf{Aut}(Z_{2^n})$ are of the form $\alpha^i\beta^j$, where $\alpha(x) = x^{5^{2^{n-3}}}$ and $\beta(x) = x^{-1}$. There are 4 of these. $\square$

Now to the main result. Write $Z_2 = \langle x \rangle$ and $Z_8 = \langle y \rangle$. The distinct homomorphisms $\varphi : Z_2 \rightarrow \mathsf{Aut}(Z_8)$ are given by $\varphi_1(x)(y) = y$, $\varphi_2(x)(y) = y^7$, $\varphi_3(x)(y) = y^5$, and $\varphi_4(x)(y) = y^3$.

Clearly $Z_8 \rtimes_{\varphi_1} Z_2 \cong Z_8 \times Z_2$.

By Example 1 in the text, $Z_8 \rtimes_{\varphi_2} Z_2 \cong D_{16}$.

We now find presentations for the remaining two groups.

$G_3 = Z_8 \rtimes_{\varphi_3} Z_2$ is generated by $x$ and $y$. We can see that $xyx^{-1} = \varphi_3(x)(y) = y^5$; so $G_3$ has the presentation $\langle x,y \ |\ x^2 = y^8 = 1, xy = y^5x \rangle$. This is precisely the defining presentation of $M$.

Similarly, $G_4 = Z_8 \rtimes_{\varphi_4} Z_2$ is generated by $x$ and $y$, and $xyx^{-1} = \varphi_4(x)(y) = y^3$. This is precisely the defining presentation for $QD_{16}$.

### Classify the groups of order 60

This exercise classifies the groups of order 60. (There are thirteen isomorphism types.)

Let $G$ be a group of order 60, let $P \leq G$ be a Sylow 5-subgroup, and let $Q \leq G$ be a Sylow 3-subgroup.

1. Prove that if $P$ is not normal in $G$, then $G \cong A_5$.
2. Prove that if $P \leq G$ is normal but that $Q \leq G$ is not normal, then $G \cong A_4 \times Z_5$. [Hint: Show that $P \leq Z(G)$, $G/P \cong A_4$, a Sylow 2-subgroup $T \leq G$ is normal, and $TQ \cong A_4$.]
3. Prove that if both $P$ and $Q$ are normal in $G$ then $G \cong Z_{15} \rtimes T$ where $T = Z_4$ or $T = Z_2^2$. Show in this case that there are six isomorphism types when $T$ is cyclic (one abelian) and five isoorphism types when $T$ is the Klein 4-group (one abelian). [Hint: Use the same ideas as in the classifications of groups of orders 20 and 30.]

[Disclaimer: I looked at parts of Alfonso Gracia-Saz’s notes for ideas while solving this problem.]

1. If $P$ is not normal, then $G$ contains at least two Sylow 5-subgroups. By Proposition 4.21, $G$ is simple. By Proposition 4.23, $G \cong A_5$.
2. Suppose $P$ is normal in $G$ and $Q$ is not normal in $G$. Since $P$ is normal in $G$, $PQ \leq G$ is a subgroup. Moreover, since $P \cap Q = 1$ by Lagrange, $|PQ| = 15$; there is only one group of order 15 up to isomorphism, so that $PQ \cong Z_{15}$. Note that $Z_{15}$ has a unique (hence normal) subgroup of order 3- namely $Q$. Since $Q \leq PQ$ is normal, we have $PQ \leq N_G(Q)$. Thus 15 divides $|N_G(Q)|$. Now $N_G(Q)/C_G(Q) \leq \mathsf{Aut}(Q) \cong Z_2$; thus 15 also divides $|C_G(Q)|$. By Cauchy, $C_G(Q)$ contains an element of order 5, and thus contains $P$. Then $Q \leq C_G(P)$. Now recall that $n_3(G) = [G : N_G(Q)]$ $= 60/(15k)$ $= 4/k$ for some integer $k$. By Sylow’s Theorem, $n_3(G) \in \{ 1,4,10 \}$, and in fact, $n_3(G) = 4$. Let $Q_1$, $Q_2$, $Q_3$, and $Q_4$ be the Sylow 3-subgroups of $G$.

Recall that $P \cap Q_i = 1$ and $P \leq G$ is normal, so that $PQ_i \leq G$ and $PQ_i \cong Z_{15}$. Now $P \leq PQ_i \cap PQ_j$ for all $i \neq j$; since each $PQ_i$ contains a unique Sylow 3-subgroup, if $|PQ_i \cap PQ_j| = 15$ then $Q_i = Q_j$, a contradiction when $i \neq j$. Thus $|PQ_i \cap PQ_j| = 5$ for all distinct $i$ and $j$, so that $PQ_i \cap PQ_j = P$. Note that each $PQ_i$ contains $\varphi(15) = \varphi(3)\varphi(5)$ $= 2 \cdot 4 = 8$ elements of order 15 (where $\varphi$ denotes the Euler totient). Moreover, if $|x| = 15$, then $P \leq \langle x \rangle$ and $Q_i \leq \langle x \rangle$ for some $i$, so that $PQ_i \leq \langle P,Q_i \rangle \leq \langle x \rangle$. Counting elements, we have $\langle x \rangle = PQ_i$. Thus every element of order 15 in $G$ is in some $PQ_i$, no element is in more than one $PQ_i$, and each $PQ_i$ contains precisely 8 elements of order 15. Thus $G$ has exactly $8 \cdot 4 = 32$ elements of order 15.

Now let $x \in G$. Recall that $Q_i \leq C_G(P)$ for some $i$. Now $xQ_ix^{-1} \leq x C_G(P) x^{-1}$ $= C_G(xPx^{-1})$ $= C_G(P)$, since $P$ is normal in $G$. By Sylow’s Theorem, all the $Q_j$ are conjugate to $Q_i$. Thus we have $PQ_i \leq C_G(P)$ for all $i$. In particular, $C_G(P)$ contains at least 32 elements; by Lagrange, $|C_G(P)| = 60$, and we have $P \leq Z(G)$.

The Sylow 3-subgroups of $G$ intersect trivially, so that every element of order 3 is in exactly one Sylow 3-subgroup. Each such subgroup contains 2 elements of order 3; thus $G$ contains exactly $2 \cdot 4 = 8$ elements of order 3. Let $X$ denote the set of all elements of order 3 or 15 in $G$; note that $|X| = 40$, so that $|G \setminus X| = 20$. Now let $T$ be a Sylow 2-subgroup of $G$. Since $P \leq G$ is normal, $PT \leq G$ is a subgroup. Since $P \cap T = 1$, $|PT| = 20$. No element of $PT$ has order 3 or 15 by Lagrange, so that $PT \subseteq G \setminus X$. Counting elements, we see that $PT = G \setminus X$. Moreover, if $H \leq G$ is a subgroup of order 20, then $H \leq PT$, so that $H = PT$. Thus, $PT$ is the unique subgroup of order 20 in $G$; thus $PT$ is characteristic, hence normal, in $G$. Note for future reference that every element of $G$ has order 1, 2, 3, 4, 5, 10, or 15.

Now every Sylow 2-subgroup of $G$ is contained in $PT$, and in fact $n_2(G) = n_2(PT)$ since the Sylow 2-subgroups of both groups have order 4. By Sylow’s Theorem, $n_2(PT) = n_2(G) \in \{1,5\}$. Recall that $n_2(G) = [G : N_G(T)]$. Since $P,T \leq N_G(T)$, we have $60/(20k) = 3/k \in \{1,5\}$, where $k$ divides $|N_G(T)|$. Thus $k = 3$, so that $N_G(T)$ contains an element of order 3 by Cauchy. Now some Sylow 3-subgroup of $G$ is contained in $PT$; without loss of generality, say $Q_1 \leq PT$. Now $N_G(T) = G$, since $P,Q_1,T \leq N_G(T)$, so that $T \leq G$ is normal. Thus $T \leq Q_1T$ is normal. Note that every element of $Q_1T \leq G$ has order 1, 2, 3, or 4; there is one identity, and there are 3 elements of order 2 or 4; the remaining elements have order 3. Thus $Q_i \leq Q_1T$ for all $i$. Now let $x \in G$. Now $Q_i, T \leq Q_1T$ for each $i$, and $T$ is normal in $G$, so that $Q_iT \leq Q_1T$ is a subgroup. Counting elements we see that $Q_iT = Q_1T$ for each $i$. Now let $x \in G$; we have $x Q_1T x^{-1} = x Q_1 x^{-1} x T x^{-1} = Q_iT = Q_1T$. Thus $Q_1T \leq G$ is normal. Finally, note that $n_3(Q_1T) = 4$ and $Q_1T$ is a group of order 12; every group of order 12 has a unique Sylow 3-subgroup except A_4$. Thus $Q_1T \cong A_4$. Finally, since $P, Q_1T \leq G$ are normal and intersect trivially, we have $G \cong P \times Q_1T$. That is, $G \cong Z_5 \times A_4$. 3. Before we proceed, we prove some technical lemmas. Lemma 1: Fix integers $m,n \geq 1$ such that, for all positive integers $a,b$, $m+a4^n \equiv a+m4^b$ (mod 5). Then $m \equiv 0$ (mod 5) and $n \equiv 0$ (mod 2). Proof: Suppose $m \not\equiv 0$ (mod 5), and let $a = 5$. Then $m \equiv m4^b$ (mod 5), and since $m$ is invertible mod 5, $4^b \equiv 1$ (mod 5)$ for all positive integers $b$. This is absurd; consider $b = 1$. Thus $m \equiv 0$ (mod 5). Now let $a \not\equiv 0$ (mod 5); we have 4^n \equiv 1$(mod 5); since $4^2 \equiv 1$ (mod 5), we have $n \equiv 0$ (mod 2). $\square$ Lemma 2: Fix integers $m,n \geq 1$ such that, for all positive integers $a,b$, $m+a2^n \equiv a+m2^b$ (mod 3). Then $m \equiv 0$ (mod 3) and $n \equiv 0$ (mod 2). Proof: Suppose $m \not\equiv 0$ (mod 3), and let $a = 3$. Then $m \equiv m2^b$, so $2^b \equiv 1$ (mod 3) for all $b$; this is absurd- consider $b = 1$. Thus $m \equiv 0$ (mod 3). Now we have $a \equiv a2^n$ (mod 3); let $a \not\equiv 0$ (mod 3). Then $2^n \equiv 1$ (mod 3). Thus $n \equiv 0$ (mod 2). $\square$ Lemma 3: Fix integers $m,n \geq 1$ such that, for all positive integers $a,b$, $m+a2^n \equiv a+m2^b$ (mod 5). Then $m \equiv 0$ (mod 5) and $n \equiv 0$ (mod 4). Proof: Suppose $m \not\equiv 0$ (mod 5), and let $a = 5$. Then $m \equiv m2^b$ (mod 5), so that $2^b \equiv 1$ (mod 5) for all positive $b$; this is absurd (consider $b = 1$) so that $m \equiv 0$ (mod 5). Thus $a \equiv a2^n$ (mod 5). If $a \not\equiv 0$ (mod 5), we have $2^n \equiv 1$ (mod 5). Thus $2 \equiv 0$ (mod 4). $\square$ Suppose now that $P,Q \leq G$ are normal and let $T$ be a Sylow 2-subgroup of $G$. As above, $PQ \cong Z_{15}$, and $PQ \leq G$ is normal. Moreover, $PQT = G$. By the recognition theorem for semidirect products, $G \cong PQ \rtimes_\varphi T$, where $\varphi : T \rightarrow \mathsf{Aut}(PQ)$. Evidently, classifying the groups of order 60 which have normal Sylow 5- and Sylow 3-subgroups is equivalent to describing the possible semidirect products $Z_{15} \rtimes T$, where $|T| = 4$. We carry out this classification now. Let $H = Z_5 \times Z_3 = \langle y \rangle \times \langle z \rangle$. Note that $Z_{15} = AB$, where $A \cong Z_5$ and $B \cong Z_3$. Moreover, being cyclic, every subgroup of $Z_{15}$ is characteristic. By this previous exercise, we have $\mathsf{Aut}(Z_{15}) \cong \mathsf{Aut}(Z_5) \times \mathsf{Aut}(Z_3)$ $\cong Z_4 \times Z_2$. Write $Z_4 \times Z_2 = \langle \alpha \rangle \times \langle \beta \rangle$, where $\alpha(y) = y^2$, $\alpha(z) = z$, $\beta(y) = y$, and $\beta(z) = z^2$. Suppose now that $K = Z_4 = \langle x \rangle$. There is a unique group homomorphism $K \rightarrow \mathsf{Aut}(H)$ for each element of order dividing 4 in $\mathsf{Aut}(H)$. Every element of $\mathsf{Aut}(H)$ has order dividing 4, so there are 8 distinct homomorphisms. 1. If $\varphi_1(x) = (1,1)$, then $G_1 = H \rtimes_{\varphi_1} K \cong Z_5 \times Z_3 \times Z_4$. 2. If $\varphi_2(x) = (\alpha,1)$, then $G_2 = H \rtimes_{\varphi_2} K$ is indeed a nonabelian group of order 60. Note that if $\varphi_3(x) = (\alpha^3,1)$, then $\varphi_3 = \varphi_2 \circ \theta$, where $\theta(\alpha) = \alpha^3$ and $\theta(\beta) = \beta$. By a lemma to a previous exercise, $H \rtimes_{\varphi_3} K \cong H \rtimes_{\varphi_2} K$. Moreover, note that $\mathsf{ker}\ \varphi_2 = 1$. 3. If $\varphi_4(x) = (\alpha,\beta)$, then $G_3 = H \rtimes_{\varphi_4} K$ is a nonabelian group of order 60. Note that if $\varphi_5(x) = (\alpha^3,\beta)$, then $\varphi_5 = \varphi_4 \circ \theta$, where $\theta(\alpha) = \alpha^3$ and $\theta(\beta) = \beta$. By a lemma to a previous exercise, $H \rtimes_{\varphi_5} K \cong H \rtimes_{\varphi_4} K$. Moreover, note that $\mathsf{ker}\ \varphi_4 = 1$. 4. If $\varphi_6(x) = (\alpha^2,1)$, then $G_4 = H \rtimes_{\varphi_6} K$ is a nonabelian group of order 60. Moreover, note that $\mathsf{ker}\ \varphi_6 = \langle x^2 \rangle \cong Z_2$. 5. If $\varphi_7(x) = (1,\beta)$, then $G_5 = H \rtimes_{\varphi_7} K$ is a nonabelian group of order 60. Moreover, note that $\mathsf{ker}\ \varphi_7 = \langle x^2 \rangle \cong Z_2$. 6. If $\varphi_8(x) = (\alpha^2,\beta)$, then $G_6 = H \rtimes_{\varphi_8} K$ is a nonabelian group of order 60. Moreover, note that $\mathsf{ker}\ \varphi_8 = \langle x^2 \rangle \cong Z_2$. We claim that these six groups are pairwise distinct. 1. Note that $G_1$ is abelian and thus distinct from the remaining five. 2. Note that $G_2$ and $G_3$ are distinct from $G_4$, $G_5$, and $G_6$ because $\varphi_2$ and $\varphi_4$ have trivial kernels while $\varphi_6$, $\varphi_7$, and $\varphi_8$ do not. We now find presentations for $G_2$ and $G_3$. $G_2$ is generated by $y$, $z$, and $x$, and $yz = zy$. Now $xy = (\varphi_2(x)(y),x)$ $y^2x$ and $xz = (\varphi_2(x)(z),x)$ $= zx$. Thus $G_2$ has the presentation $\langle x,y,z \ |\ x^4 = y^5 = z^3 = 1,$ $yz = zy, zx = xz, xy = y^2x \rangle$. $G_3$ is generated by $y$, $z$, and $x$, and $yz = zy$. Now $xy = (\varphi_4(x)(y),x)$ $= y^2x$ and $xz = z^2x$. Thus $G_3$ has the presentation $\langle x,y,z \ |\ x^4 = y^5 = z^3 = 1,$ $yz = zy, xy = y^2x, xz = z^2x \rangle$. Now we compute the centers of $G_2$ and $G_3$. Every element of $G_2$ can be written in the form $y^iz^jx^k$, where $0 < i \leq 5$, $0 < j \leq 3$, and $0 < k \leq 4$. There are 60 such expressions, and we know by other means that $|G_2| = 60$. Thus every element can be written in this way uniquely. Let $s = y^az^bx^c \in Z(G_2)$ and let $y^iz^jx^k \in G_2$. Then, using a lemma to this previous exercise, $y^az^bx^cy^iz^jx^k = y^az^by^{i2^c}x^cz^jx^k$ $= y^ay^{i2^c}z^bz^jx^cx^k$ $= y^{a+i2^c}z^{b+j}x^{c+k}$ is equal to $y^iz^jx^ky^az^bx^c = y^iz^jy^{a2^k}x^kz^bx^c$ $= y^{i+a2^k}z^{j+b}x^{k+c}$. Comparing exponents, we have $a+i2^c \equiv i+a2^k$ (mod 5). By Lemma 2, we have $a \equiv 0$ (mod 5) and $c \equiv 0$ (mod 4). Thus $s \in \langle z \rangle$. Clearly $z \in Z(G_2)$, so that $Z(G_2) = \langle z \rangle \cong Z_3$. Similarly, every element of $G_3$ can be written uniquely in the form $y^iz^jx^k$, where $0 < i \leq 5$, $0 < j \leq 3$, and $0 < k \leq 4$. Let $s = y^az^bx^c \in Z(G_3)$ and let $y^iz^jx^k \in G_3$. Then $y^az^bx^cy^iz^jx^k = y^az^by^{i2^c}x^cz^jx^k$ $= y^ay^{i2^c}z^bz^{j2^c}x^cx^k$ $= y^{a+i2^c}z^{b+j2^c}x^{c+k}$ is equal to $y^iz^jx^ky^az^bx^c = y^iz^jy^{a2^k}x^kz^bx^c$ $= y^{i+a2^k}z^{j+b2^k}x^{k+c}$. Comparing exponents, we have $i+a2^k \equiv a+i2^c$ (mod 5) and $j+b2^k \equiv b+j2^c$ (mod 3). By Lemmas 2 and 3, we have $a \equiv 0$ (mod 5), $b \equiv 0$ (mod 3), and $c \equiv 0$ (mod 4). Thus $s = 1$, and we have $Z(G_3) = 1$. In particular, $G_2 \not\cong G_3$ since their centers are not isomorphic. 3. Next, we find presentations for $G_4$, $G_5$, and $G_6$. $G_4$ is generated by $y$, $z$, and $x$, and $yz = zy$. Now $xy = (1,x)(y,1)$ $= (\varphi_6(x)(y),x)$ $= (\alpha^2(y),x)$ $= (y^4,x)$ $= y^4x$. Similarly, $xz = (\varphi_6(x)(z),x)$ $= (z,x)$ $= zx$. Thus $G_4$ has the presentation $\langle x,y,z \ |\ x^4 = y^5 = z^3 = 1,$ $yz = zy, xz = zx, xy = y^4x \rangle$. $G_5$ is generated by $y$, $z$, and $x$, and $yz = zy$. Now $xy = (\varphi_7(x)(y),x)$ $= yx$ and $xz = z^2x$. Thus $G_5$ has the presentation $\langle x,y,z \ |\ x^4 = y^5 = z^3 = 1,$ $yz = zy, xy = yx, xz = z^2x \rangle$. $G_6$ is generated by $y$, $z$, and $x$, and $yz = zy$. Now $xy = (\varphi_8(x)(y),x)$ $= y^4x$ and $xz = (\varphi_8(x)(z),x)$ $= z^2x$. Thus $G_6$ has the presentation $\langle x,y,z \ |\ x^4 = y^5 = z^3 = 1,$ $yz = zy, xy = y^4x, xz = z^2x \rangle$. Now we compute the centers of $G_4$, $G_5$, and $G_6$. Note that every element of $G_4$ can be written in the form $y^iz^jx^k$, where $0 < i \leq 5$, $0 < j \leq 3$, and $0 < k \leq 4$. There are 60 such expressions, and we know by other means that $|G_4| = 60$. Thus this representation of an element in $G_4$ is unique. Now let $s = y^az^bx^c \in Z(G_4)$, and let $y^iz^jx^k \in G_4$. Then, using a lemma to this previous exercise, $y^az^bx^cy^iz^jx^k = y^az^by^{i4^c}x^cz^jx^k$ $= y^ay^{i4^c}z^bz^jx^cx^k$ $= y^{a+i4^c}z^{b+j}x^{c+k}$ is equal to $y^iz^jx^ky^az^bx^c = y^iz^jy^{a4^k}x^kz^bx^c$ $= y^{i + a4^k}z^{j+b}x^{k+c}$. Comparing exponents, we have $a+i4^c \equiv i + a4^k$ (mod 5). By Lemma 1, this implies that $a \equiv 0$ (mod 5) and $c \equiv 0$ (mod 2). Thus $s = z^b$ or $s = z^bx^2$. Hence $s \in \langle z, x^2 \rangle$. Moreover, it is clear that $z \in Z(G_4)$ and $x^2 \in Z(G_4)$ since $x^2y = xy^4x$ $= y^{16}x^2$ $= yx^2$. Thus $Z(G_4) = \langle z, x^2 \rangle \cong Z_6$. Similarly, every element of $G_5$ can be written uniquely in the form $y^iz^jx^k$. Let $s = y^az^bx^c \in Z(G_5)$ and let $y^iz^jx^k \in G_5$. Then $y^az^bx^cy^iz^jx^k = y^az^by^ix^cz^jx^k$ $= y^ay^iz^bz^{j2^c}x^cx^k$ $= y^{a+i}z^{b+j2^c}x^{c+k}$ is equal to $y^iz^jx^ky^az^bx^c = y^iz^jy^ax^kz^bx^c$ $y^iy^az^jz^{b2^k}x^kx^c$ $y^{i+a}z^{j + b2^k}x^{k+c}$. Comparing exponents, we have $b \equiv 0$ (mod 3) and $c \equiv 0$ (mod 2) by Lemma 2. Thus $s = y^b$ or $s = y^bx^2$, so that $s \in \langle y,x^2 \rangle$. We can see that $y \in Z(G_5)$ and $x^2 \in Z(G_5)$ since $x^2z = xz^2x$ $= z^4x^2$ $= zx^2$. Thus $Z(G_5) = \langle y, x^2 \rangle \cong Z_{10}$. Similarly, every element of $G_6$ can be written uniquely in the form $y^iz^jx^k$. Let $s = y^az^bx^c \in Z(G_6)$ and let $y^iz^jx^k \in G_6$. Then $y^az^bx^cy^iz^jx^k = y^az^by^{i4^c}x^cz^jx^k$ $= y^ay^{i4^c}z^bz^{j2^c}x^cx^k$ $= y^{a+i4^c}z^{b+j2^c}x^{c+k}$ is equal to $y^iz^jx^ky^az^bx^c = y^iz^jy^{a4^k}x^kz^bx^c$ $y^iy^{a4^k}z^jz^{b2^k}x^kx^c$ $y^{i + a4^k}z^{j + b2^k}x^{k+c}$. Comparing exponents, we have $a+i4^c \equiv i + a4^k$ (mod 5) and $b+j2^c \equiv j + b2^k$ (mod 3). By Lemmas 1 and 2, we have $a \equiv 0$ (mod 5), $b \equiv 0$ (mod 3), and $c \equiv 0$ (mod 2). Thus $s \in \langle x^2 \rangle$. Moreover, note that $x^2y = xy^4x$ $= y^{16}x^2$ $= yx^2$ and $x^2z = xz^2x$ $= z^4x^2$ $= zx^2$, so that $x^2 \in Z(G_6)$. Thus $Z(G_6) = \langle x^2 \rangle \cong Z_2$. In particular, $G_4$, $G_5$, and $G_6$ are mutually nonisomorphic because their centers are not isomorphic. 4. Thus there are precisely 6 nonisomorphic groups of order 60 which have normal Sylow 5- and Sylow 3-subgroups and whose Sylow 2-subgroups are cyclic. Suppose now that $K = Z_2^2 = \langle v \rangle \times \langle w \rangle$. Every homomorphism $\psi : K \rightarrow \mathsf{Aut}(H)$ is uniquely determined by $\psi(v)$ and $\psi(w)$, and such an assignment determines a homomorphism precisely when $|\psi(v)|$ and $|\psi(w)|$ divide 2. We can easily see that the number of elements of order dividing 2 in $Z_4 \times Z_2$ is four: $(\alpha^2,1)$, $(1,\beta)$, $(\alpha^2,\beta)$, and $(1,1)$. This gives us 16 distinct homomorphisms $\psi : Z_2^2 \rightarrow Z_4 \times Z_2$. We summarize these mappings in the following table; rows and columns are indexed by $\psi(w)$ and $\psi(v)$, respectively.  $\psi_i(v)$ $1$ $\alpha^2$ $\beta$ $\alpha^2\beta$ $\psi_i(w)$ $1$ $\psi_1$ $\psi_2$ $\psi_3$ $\psi_4$ $\alpha^2$ $\psi_5$ $\psi_6$ $\psi_7$ $\psi_8$ $\beta$ $\psi_9$ $\psi_{10}$ $\psi_{11}$ $\psi_{12}$ $\alpha^2\beta$ $\psi_{13}$ $\psi_{14}$ $\psi_{15}$ $\psi_{16}$ Also, define the following mappings $\theta_i : Z_2^2 \rightarrow Z_2^2$. 1. $\theta_1(v) = w$ and $\theta_1(w) = v$ 2. $\theta_2(v) = v$ and $\theta_2(w) = vw$ 3. $\theta_3(v) = vw$ and $\theta_3(w) = w$ 4. $\theta_4(v) = w$ and $\theta_4(w) = vw$ 5. $\theta_5(v) = vw$ and $\theta_5(w) = v$ Note the following. 1. $\psi_5 = \psi_2 \circ \theta_1$ and $\psi_6 = \psi_2 \circ \theta_2$ 2. $\psi_9 = \psi_3 \circ \theta_1$ and $\psi_{11} = \psi_3 \circ \theta_2$ 3. $\psi_{13} = \psi_4 \circ \theta_1$ and $\psi_{16} = \psi_4 \circ \theta_2$ 4. $\psi_{7} = \psi_{10} \circ \theta_1$, $\psi_{14} = \psi_{10} \circ \theta_2$, $\psi_{12} = \psi_{10} \circ \theta_3$, $\psi_{15} = \psi_{10} \circ \theta_4$, and $\psi_{8} = \psi_{10} \circ \theta_5$ It is easy to see that each $\theta_i$ is an isomorphism; thus, by a lemma to a previous exercise, the semidirect products formed by any two $\psi_i$ which appear in an equation $\psi_i = \psi_j \circ \theta_k$ are isomorphic. This gives us at most five distinct groups, which we name as follows. 1. $G_7 = H \varphi_{\psi_1} K$ 2. $G_8 = H \varphi_{\psi_2} K$ 3. $G_9 = H \varphi_{\psi_3} K$ 4. $G_{10} = H \varphi_{\psi_4} K$ 5. $G_{11} = H \varphi_{\psi_{10}} K$ We claim that these five are indeed distinct. Note that $G_7 \cong Z_5 \times Z_3 \times Z_2^2$ is abelian, while the other four are not; thus $G_7$ is distinct from the other candidates. We can easily see that $\mathsf{ker}\ \psi_2 \cong \mathsf{ker}\ \psi_3 \cong \mathsf{ker}\ \psi_4 \cong Z_2^2$ while $\mathsf{ker}\ \psi_{10} \cong Z_2$; thus $G_{11}$ is distinct from $G_8$, $G_9$, and $G_{10}$. Next we compute presentations for $G_8$, $G_9$, and $G_{10}$. $G_8$ is generated by $y$, $z$, $v$, and $w$, and we have $yz = zy$ and $vw = wv$. Now $vy = (1,v)(y,1)$ $= (\psi_2(v)(y),v)$ $= (\alpha^2(y),v)$ $= y^4v$, $wy = (\psi_2(w)(y),w)$ $= yw$, $vz = (\psi_2(v)(z),v)$ $= (\alpha^2(z),v)$ $= zv$, and $wz = (\psi_2(w)(z),w)$ $= zw$. Thus $G_8$ has the presentation $\langle y,z,v,w \ |\ y^5 = z^3 = v^2 = w^2 = 1,$ $yz = zy, vw = wv,$ $vy = y^4v$, $wy = yw$, $vz = zv$, $wz = zw \rangle$. $G_9$ is generated by $y$, $z$, $v$, and $w$, and we have $yz = zy$ and $vw = wv$. Now $vy = (\psi_3(v)(y),v)$ $= (\beta(y),v)$ $= yv$, $wy = (\psi_3(w)(y),w)$ $= yw$, $vz = (\psi_3(v)(z),v)$ $= z^2v$, and $wz = (\psi_3(w)(v),w)$ $= vw$. Thus $G_9$ has the presentation $\langle y,z,v,w \ |\ y^5 = z^3 = v^2 = w^2 = 1,$ $yz = zy, vw = wv,$ $vy = yv,$ $wy = yw,$ $vz = z^2v,$ $wv = vw \rangle$. $G_{10}$ is generated by $y$, $z$, $v$, and $w$, and we have $yz = zy$ and $vw = wv$. By a similar process, we see that $G_{10}$ has the presentation $\langle y,z,v,w \ |\ y^5 = z^3 = v^2 = w^2 = 1,$ $yz = zy, vw = wv,$ $vy = y^4v,$ $wy = yw,$ $vz = z^2v,$ $wv = vw \rangle$. Next, we compute the centers of $G_8$, $G_9$, and $G_{10}$. Every element of $G_8$ can be written in the form $y^iz^jv^kw^\ell$, where $0 < i \leq 5$, $0 < j \leq 3$, and $0 < k,\ell \leq 2$. There are 60 such forms, and we know by other means that $|G_8| = 60$. Thus every element can be written in this form uniquely. Let $s = y^az^bv^cw^d \in Z(G_8)$ and let $y^iz^jv^kw^\ell \in G_8$. Now $y^az^bv^cw^dy^iz^jv^kw^\ell = y^az^bv^cy^iw^dz^jv^kw^\ell$ $= y^az^by^{i4^c}v^cz^jw^dv^kw^\ell$ $= y^ay^{i4^c}z^bz^jv^cv^kw^dw^\ell$ $= y^{a+i4^c}z^{b+j}v^{c+k}w^{d+\ell}$ is equal to $y^{i+a4^k}z^{b+j}v^{c+k}w^{d+\ell}$. Comparing exponents, we have $a+i4^c \equiv i+a4^k$ (mod 5). By Lemma 1, this implies $a \cong 0$ (mod 5) and $c \cong 0$ (mod 2). Thus $s = z^bw^d \in \langle z,w \rangle$. We can easily see that $\langle z,w \rangle \leq Z(G_8)$, so that $Z(G_8) = \langle z,w \rangle \cong Z_6$. Every element of $G_9$ can also be written uniquely in the form $y^iz^jv^kw^\ell$, where $0 < i \leq 5$, $0 < j \leq 3$, and $0 < k,\ell \leq 2$. Let $s = y^az^bv^cw^d \in Z(G_9)$ and let $y^iz^jv^kw^\ell \in G_9$. Now $y^az^bv^cw^dy^iz^jv^kw^\ell = y^az^bv^cy^iw^dz^jv^kw^\ell$ $= y^az^by^iv^cz^jw^dv^kw^\ell$ $= y^ay^iz^bz^{j2^c}v^cv^kw^dw^\ell$ $= y^{a+i}z^{b+j2^c}v^{c+k}w^{d+\ell}$ is equal to $y^{i+a}z^{b+j2^k}v^{c+k}w^{d+\ell}$. Comparing exponents and using Lemma 2, we see that $b \equiv 0$ (mod 3) and $c \equiv 0$ (mod 2). Thus $s = y^aw^d \in \langle y,w \rangle$. Clearly $\langle y,w \rangle \leq Z(G_9)$, so that $Z(G_9) = \langle y,w \rangle \cong Z_{10}$. Every element of $G_{10}$ can also be written uniquely in the form $y^iz^jv^kw^\ell$, where $0 < i \leq 5$, $0 < j \leq 3$, and $0 < k,\ell \leq 2$. Let $s = y^az^bv^cw^d \in Z(G_{10})$ and let $y^iz^jv^kw^\ell \in G_{10}$. Now $y^az^bv^cw^dy^iz^jv^kw^\ell = y^az^bv^cy^iw^dz^jv^kw^\ell$ $= y^az^by^{i4^c}v^cz^jw^dv^kw^\ell$ $= y^ay^{i4^c}z^bz^{j2^c}v^cv^kw^dw^\ell$ $= y^{a+i4^c}z^{b+j2^c}v^{c+k}w^{d+\ell}$ is equal to $y^{i+a4^k}z^{b+j2^k}v^{c+k}w^{d+\ell}$. Comparing exponents and using Lemma 2, we see that $a \equiv 0$ (mod 5), $b \equiv 0$ (mod 3), and $c \equiv 0$ (mod 2). Thus $s = w^d \in \langle w \rangle$. Clearly $\langle w \rangle \leq Z(G_{10})$, so that $Z(G_{10}) = \langle w \rangle \cong Z_{2}$. Clearly then $G_8$, $G_9$, and $G_{10}$ are distinct because their centers are not isomorphic. This completes the classification of groups of order 60. ### Classify groups of order 4p, where p is a prime greater than 3 Classify groups of order $4p$, where $p$ is a prime greater than 3. By FTFGAG, there are two distinct groups of order $4p$: $Z_{4p}$ and $Z_{2p} \times Z_2$. Now let $G$ be a nonabelian group of order $4p$. By Sylow’s Theorem, $n_p$ is congruent to 1 mod $p$ and divides 4; since $p > 4$, we thus have $n_p = 1$. Let $H \leq G$ be the unique (hence normal) Sylow $p$-subgroup of $G$. Now let $K \leq G$ be any Sylow 2-subgroup. By Lagrange, $H \cap K = 1$, so that $G = HK$. By the recognition theorem for semidirect products, we have $G \cong H \rtimes_\varphi K$ for some homomorphism $\varphi K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying nonabelian groups of order $4p$ is equivalent to determining the nonisomorphic groups which may be constructed in this manner. To that end, let $H = Z_p = \langle y \rangle$. Now $\mathsf{Aut}(H) \cong Z_{p-1}$ is cyclic; say $\mathsf{Aut}(H) = \langle \alpha \rangle$, where $\alpha(y) = y^2$. Before proceeding, we prove the following lemma. Lemma 1: Let $p \geq 5$ be a prime, and write $Z_4 = \langle x \rangle$ and $Z_{p-1} = \langle y \rangle$. 1. If $p \equiv 3$ (mod 4), then there is a unique nontrivial group homomorphism $\varphi : Z_4 \rightarrow Z_{p-1}$, given by $\varphi(x) = y^{(p-1)/2}$. 2. If $p \equiv 1$ (mod 4), then there are exactly three distinct nontrivial group homomorphisms $\psi_1,\psi_2, \psi_3 : Z_4 \rightarrow Z_{p-1}$, given by $\psi_1(x) = y^{(p-1)/2}$, $\psi_2(x) = y^{(p-1)/4}$, and $\psi_3(x) = y^{3(p-1)/4}$. Proof: Note that an assignment $\varphi(x)$ extends to a unique homomorphism if and only if $|\varphi(x)|$ divides 4. Thus we must search for elements of order 2 and 4 in $Z_{p-1}$. 1. If $p \equiv 3$ (mod 4), then 2 divides $p-1$ but 4 does not. By Lagrange, $Z_{p-1}$ contains no elements of order 4. Now by Theorem 2.7, there is a unique subgroup (hence element) of order 2 in $Z_{p-1}$; namely $\langle y^{(p-1)/2} \rangle$. Thus $\varphi_1(x) = y^{(p-1)/2}$ is the unique nontrivial homomorphism $Z_4 \rightarrow Z_{p-1}$. 2. If $p \equiv 1$ (mod 4), then 4 divides $p-1$. By Theorem 2.7, there exist unique subgroups of order 4 and 2 in $Z_{p-1}$; namely, $\langle y^{(p-1)/4} \rangle$ and $\langle y^{(p-1)/2} \rangle$, respectively. $\square$ Let $K = Z_4 = \langle x \rangle$. If $p \equiv 3$ (mod 4), there is a unique nontrivial homomorphism $\varphi : Z_4 \rightarrow Z_{p-1}$, which gives rise to a nonabelian group of order $4p$: $Z_p \rtimes_\varphi Z_4$. If $p \equiv 1$ (mod 4), there are three distinct nontrivial homomorphisms $\psi_1,\psi_2,\psi_3 : Z_4 \rightarrow Z_{p-1}$ as described in the lemma. Note that $\mathsf{im}\ \psi_2$ and$\mathsf{im}\ \psi_3$are equal, and hence conjugate subgroups of $\mathsf{Aut}(H)$. Since $K$ is cyclic, by a previous theorem, we have $H \rtimes_{\psi_2} K \cong H \rtimes_{\psi_3} K$. On the other hand, $H \rtimes_{\psi_1} K$ and $H \rtimes_{\psi_2} K$ are distinct since $\mathsf{ker}\ \psi_1 \cong Z_2$ while $\mathsf{ker}\ \psi_2$ is trivial, using the comments in part (c) of §5.5 #7. Thus if $p \equiv 3$ (mod 4), there is a unique nonabelian group of order $4p$ which has a cyclic Sylow 2-subgroup, while if $p \equiv 1$ (mod 4), there are two distinct nonabelian groups of this type. Before we proceed, we give another lemma. Lemma 2: Let $p \geq 5$ be a prime. $Z_{p-1} = \langle y \rangle$ has a unique element of order 2; namely $y^{(p-1)/2}$. Proof: Elements and subgroups of order 2 in $Z_{p-1}$ are equivalent, and by Theorem 2.7, $Z_{p-1}$ has a unique subgroup of order 2. $\square$ Now let $K = Z_2^2 = \langle a \rangle \times \langle b \rangle$. Group homomorphisms $\psi : K \rightarrow \mathsf{Aut}(H)$ correspond precisely to the distinct assignments of $a$ and $b$ to elements of order 1 or 2 in $Z_{p-1}$. By Lemma 2, there is exactly one element of order 2 in $Z_{p-1}$, namely $\alpha^{(p-1)/2}$; thus there are four distinct homomorphisms. If $\psi_1(a) = \psi1(b) = 1$, then $\psi$ is trivial, contradicting the nonabelianicity of $G$. If $\psi_2(a) = \alpha^{(p-1)/2}$ and $\psi_2(b) = 1$, then $H \rtimes_{\psi_2} K$ is indeed a nonabelian group of order $4p$. If $\psi_3(a) = 1$ and $\psi_3(b) = \alpha^{(p-1)/2}$, then we have $\psi_3 = \psi_2 \circ \theta$ where $\theta(a) = b$ and $\theta(b) = a$. Now $\theta$ is clearly an automorphism of $K$, so that by a lemma to §5.5 #11, $H \rtimes_{\psi_3} K \cong H \rtimes_{\psi_2} K$. If $\psi_4(a) = \psi_4(b) = \alpha^{(p-1)/2}$, then we have $\psi_4 = \psi_2 \circ \theta$ where $\theta(a) = a$ and $\theta(b) = ab$. Again, this implies that $H \rtimes_{\psi_4} K \cong H \rtimes_{\psi_2} K$. Thus there is a unique nonabelian group of order $4p$ which has an elementary abelian Sylow 2-subgroup. In summary, the distinct groups of order $4p$, $p$ a prime, are as follows. We let $Z_p = \langle y \rangle$, $Z_4 = \langle x \rangle$, and $Z_2^2 = \langle a \rangle \times \langle b \rangle$. For all $p$, we have 1. $Z_{4p}$ 2. $Z_{2p} \times Z_2$ 3. $Z_p \rtimes_\varphi Z_4$, where $\varphi(x)(y) = y^{-1}$ 4. $Z_p \rtimes_\psi Z_2^2$, where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$ If $p \equiv 1$ (mod 4), we also have 1. $Z_p \rtimes_\chi Z_4$, where $\chi(x)(y) = y^{(p-1)/2}$ ### Classification of groups of order 20 Classify the groups of order 20. [Hint: There are five isomorphism types.] Note that $20 = 2^2 \cdot 5$. By FTFGAG, there are two distinct abelian groups of order 20: $Z_{20}$ and $Z_{10} \times Z_2$. Now let $G$ be a nonabelian group of order 20. By Sylow’s Theorem, $n_5 = 1$, so that $G$ has a unique (hence normal) Sylow 5-subgroup $H \cong Z_5$. Now let $K$ be any Sylow 2-subgroup of $G$. By Lagrange, we have $H \cap K = 1$, so that $G = HK$. By the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$ for some $\varphi : K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying the nonabelian groups of order 20 is equivalent to determining the nonisomorphic groups constructed in this manner. To that end, let $H = Z_5 = \langle y \rangle$. Note that $\mathsf{Aut}(H) = \langle \alpha \rangle \cong Z_4$; where $\alpha(y) = y^2$. Let $K = Z_4 = \langle x \rangle$. There are four distinct homomorphisms $K \rightarrow \mathsf{Aut}(H)$. If $\varphi_1(x) = 1$, then $\varphi_1$ is trivial; this contradicts the nonabelianicity of $G$. If $\varphi_2(x) = \alpha$, then $Z_5 \rtimes_{\varphi_2} Z_4$ is indeed a nonabelian group of order 20. If $\varphi_3(x) = \alpha^2$, then $Z_5 \rtimes_{\varphi_3} Z_4$ is indeed a nonabelian group of order 20. Moreover, since $\mathsf{ker}\ \varphi_3 \cong Z_2$ and $\mathsf{ker}\ \varphi_2 \cong 1$, $H \rtimes_{\varphi_3} K \not\cong H \rtimes_{\varphi_2} K$. If $\varphi_4(x) = \alpha^3$, then $\mathsf{im}\ \varphi_4 = \mathsf{im}\ \varphi_2$. Since $Z_4$ is cyclic, by a previous theorem, $H \rtimes_{\varphi_4} K \cong H \rtimes_{\varphi_2} K$. Thus there are two distinct groups of order 20 which have a cyclic Sylow 2-subgroup. Suppose now that $K = Z_2^2 = \langle a \rangle \times \langle b \rangle$. Again, $\psi : Z_2^2 \rightarrow Z_4$ is determined uniquely by $\psi(a)$ and $\psi(b)$, and is indeed a homomorphism provided $|\psi(a)|$ and $|\psi(b)|$ divide 2. We thus have $\psi(a), \psi(b) \in \{ 1, \alpha^2 \}$, for a total of four choices. If $\psi_1(a) = \psi_1(b) = 1$, then $\psi_1 = 1$, contradicting the nonabelianicity of $G$. If $\psi_2(a) = \alpha^2$ and $\psi_2(b) = 1$, then $Z_5 \rtimes_{\psi_2} Z_2^2$ is indeed a nonabelian group of order 20. If $\psi_3(a) = 1$ and $\psi_3(b) = \alpha^2$, then $\varphi_3 = \varphi_2 \circ \theta$, where $\theta(a) = b$ and $\theta(b) = a$. Clearly $\theta$ is an automorphism of $Z_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_3} K \cong H \rtimes_{\psi_2} K$. If $\psi_4(a) = \alpha^2$ and $\psi_4(b) = \alpha^2$, then $\varphi_4 = \varphi_2 \circ \theta$, where $\theta(a) = a$ and $\theta(b) = ab$. Clearly $\theta$ is an automorphism of $Z_2^2$. By a lemma to a previous theorem, we have $H \rtimes_{\psi_4} K \cong H \rtimes_{\psi_2} K$. Thus there is a unique nonabelian group of order 20 which has an elementary abelian Sylow 2-subgroup. In summary, the distinct groups of order 20 are as follows. We let $Z_5 = \langle y \rangle$, $Z_4 = \langle x \rangle$, and $Z_2^2 = \langle a \rangle \times \langle b \rangle$. 1. $Z_{20}$ 2. $Z_{10} \times Z_2$ 3. $Z_5 \rtimes_{\varphi_3} Z_4$, where $\varphi_3(x)(y) = y^{-1}$. 4. $Z_5 \rtimes_{\varphi_2} Z_4$, where $\varphi_2(x)(y) = y^2$ 5. $Z_5 \rtimes_\psi Z_2^2$, where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$. ### Classification of groups of order 28 Classify the finite groups of order 28. We begin with a lemma. Lemma 1: Let $H$ and $K$ be groups, $\varphi, \psi : K \rightarrow \mathsf{Aut}(H)$ group homomorphisms, and $\theta : K \rightarrow K$ an automorphism. If $\psi = \varphi \circ \theta$, then the mapping $\Phi : H \rtimes_\varphi K \rightarrow H \rtimes_\psi K$ given by $\Phi((h,k)) = (h,\theta^{-1}(k))$ is an isomorphism. Proof: (Homomorphism) Let $(h_1,k_1), (h_2,k_2) \in H \rtimes_\varphi K$. Then $\Phi((h_1,k_1)(h_2,k_2)) = \Phi((h_1 \varphi(k_1)(h_2), k_1k_2))$ $= (h_1 \varphi(k_1)(h_2), \theta^{-1}(k_1k_2))$ $= (h_1 \varphi(\theta(\theta^{-1}(k_1)))(h_2), \theta^{-1}(k_1) \theta^{-1}(k_2))$ $= (h_1 (\varphi \circ \theta)(\theta^{-1}(k_1))(h_2), \theta^{-1}(k_1) \theta^{-1}(k_2))$ $= (h_1 \psi(\theta^{-1}(k_1))(h_2), \theta^{-1}(k_1) \theta^{-1}(k_2))$ $= (h_1, \theta^{-1}(k_1))(h_2, \theta^{-1}(k_2))$ $\Phi((h_1,k_1)) \Phi((h_2,k_2))$. (Injective) If $\Phi((h_1,k_1)) = \Phi((h_2,k_2))$, then $(h_1,\theta^{-1}(k_1)) = (h_2,\theta^{-1}(k_2))$, so that $h_1 = h_2$ and $k_1 = k_2$. (Surjective) If $(h,k) \in H \rtimes_\psi K$, then $(h,k) = \Phi((h, \theta(k)))$. $\square$ Now to the main result. Note that $28 = 2^2 \cdot 7$. By FTFGAG, there are two distinct abelian groups of order 28: $Z_{28}$ and $Z_{14} \times Z_2$. Suppose now that $G$ is a nonabelian group of order 28. By Sylow’s Theorem, we have $n_7 = 1$; thus $G$ has a unique (hence normal) Sylow 7-subgroup $H \cong Z_7$. Let $K \leq G$ be a Sylow 2-subgroup. By Lagrange, we have $H \cap K = 1$; thus $G = HK$, and by the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$ for some $\varphi : K \rightarrow \mathsf{Aut}(H)$. Evidently, classifying the nonabelian groups of order 28 is equivalent to classifying the distinct groups constructed in this way. To that end, let $H = Z_7 = \langle y \rangle$. Then $\mathsf{Aut}(Z_7) \cong Z_6 = \langle \alpha \rangle$, where $\alpha(x) = x^2$. Suppose $K = Z_4 = \langle x \rangle$. Now every homomorphism $\varphi : Z_4 \rightarrow Z_6$ is determined by $\varphi(x)$, and moreover, provided $\varphi(x)^4 = 1$, any map thus defined is indeed a homomorphism. Thus we have two choices for $\varphi(x)$: $1$ and $\alpha^3$. If $\varphi(x) = 1$, then $\varphi$ is trivial, contradicting the nonabelianicity of $G$. Thus $\varphi(x) = \alpha^3$ is the only group homomorphism $K \rightarrow \mathsf{Aut}(H)$ which gives rise to a nonabelian group $Z_7 \rtimes_\varphi Z_4$ of order 28. Suppose now that $K = Z_2^2 = \langle a \rangle \times \langle b \rangle$. Every homomorphism $\varphi : K \rightarrow \mathsf{Aut}(H)$ is determined uniquely by $\varphi(a)$ and $\varphi(b)$, and provided $\varphi(a)^2 = \varphi(b)^2 = 1$, every mapping thus defined is a homomorphism. Thus we have $\varphi(a), \varphi(b) \in \{1, \alpha^3 \}$, for a total of four choices. If $\varphi_1(a) = \varphi_1(b) = 1$, then $\varphi$ is trivial, contradicting the nonabelianicity of $G$. If $\varphi_2(a) = \alpha^3$ and $\varphi_2(b) = 1$, then $Z_7 \rtimes_{\varphi_2} Z_2^2$ is indeed a nonabelian group of order 28. If $\varphi_3(a) = 1$ and $\varphi_3(b) = \alpha^3$, then we have $\varphi_3 = \varphi_2 \circ \theta$, where $\theta(a) = b$ and $\theta(b) = a$. By the Lemma, $H \rtimes_{\varphi_3} K \cong H \rtimes_{\varphi_2} K$. Likewise, if $\varphi_4(a) = \alpha^3$ and $\varphi_4(b) = \alpha^3$, we do not get an essentially new group by the lemma, via the map $\theta(a) = a$ and $\theta(b) = ab$. Thus there is a unique nonabelian group of order 28 whose Sylow 2-subgroups are not cyclic. Thus, the distinct groups of order 28 are as follows. In all cases, $Z_7 = y$, $Z_4 = \langle x \rangle$, and $Z_2^2 = \langle a \rangle \times \langle b \rangle$. 1. $Z_{28}$ 2. $Z_{14} \times Z_2$ 3. $Z_7 \rtimes_\varphi Z_4$ where $\varphi(x)(y) = y^{-1}$ 4. $Z_7 \rtimes_\psi Z_2^2$ where $\psi(a)(y) = y^{-1}$ and $\psi(b)(y) = y$ ### Classify the groups of order 147 This exercise classifies the groups of order 147. (There are six isomorphism types.) 1. Prove that there are two abelian groups of order 147. 2. Prove that every group of order 147 has a normal Sylow 7-subgroup. 3. Prove that there is a unique nonabelian group of order 147 whose Sylow 7-subgroup is cyclic. 4. Let $t_1 = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$ and $t_2 = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$ be elements of $GL_2(\mathbb{F}_7)$. Prove that $P = \langle t_1, t_2 \rangle$ is a Sylow 3-subgroup of $GL_2(\mathbb{F}_7)$ and that $P \cong Z_3^2$. Deduce that every subgroup of order 3 in $GL_2(\mathbb{F}_7)$ is conjugate to some subgroup of $P$. 5. By Example 3 in Section 5.1 of the text, the group $P$ has 4 subgroups of order 3 as follows: $P_1 = \langle t_1 = \alpha_1 \rangle$, $P_2 = \langle t_2 = \alpha_2 \rangle$, $P_3 = \langle t_1t_2 = \alpha_3 \rangle$, and $P_4 = \langle t_1t_2^2 = \alpha_4 \rangle$. For each $i$, let $G_i = Z_7^2 \rtimes_{\varphi_i} Z_3$, where $Z_3 = \langle x \rangle$ and $\varphi : Z_3 \rightarrow GL_2(\mathbb{F}_7)$ is given by $\varphi(x) = \alpha_i$. For each $i$ give a presentation for $G_i$; deduce that $G_1 \cong G_2$. 6. Prove that $G_1$ is not isomorphic to either of $G_3$ and $G_4$. [Hint: Show that the center of $G_1$ has order 7 while the centers of $G_3$ and $G_4$ are trivial.] 7. Prove that $G_3$ is not isomorphic to $G_4$. [Hint: Show that every subgroup of order 7 in $G_3$ is normal there but that $G_4$ contains order 7 subgroups which are not normal.] 8. Classify the groups of order 147 by showing that the six nonisomorphic groups described above are all the groups of order 147. The classification of groups of order $pq^2$ where $p$ and $q$ are primes, $p < q$, and $p|q-1$ is quite similar. 1. Note that $147 = 3 \cdot 7^2$. By FTFGAG, the abelian groups of order 147 are (up to isomorphism) $Z_{147}$ and $Z_{21} \times Z_7$. 2. By Sylow’s Theorem, the number $n_7$ of Sylow 7-subgroups divides 3 and is congruent to 1 mod 7; thus $n_7 = 1$. Thus the Sylow 7-subgroup of a group of order 147 is unique, hence normal. 3. Let $G$ be a nonabelian group of order 147. Let $H \leq G$ be the (unique hence normal) Sylow 7-subgroup of $G$, and suppose $H \cong Z_{49}$. Now let $K = \langle x \rangle \cong Z_3$ be any Sylow 3-subgroup of $G$. By Lagrange, $H \cap K = 1$, so that $HK = G$. By the recognition theorem for semidirect products, we have $G \cong Z_{49} \rtimes_\varphi Z_3$ for some group homomorphism $\varphi : Z_3 \rightarrow \mathsf{Aut}(Z_{49})$. Note that $\mathsf{Aut}(H) \cong Z_{42}$ and that $42 = 2 \cdot 3 \cdot 7$. By Cauchy’s Theorem, $\mathsf{Aut}(H)$ contains an element $\alpha$ of order 3. Moreover, the Sylow 3-subgroups of $\mathsf{Aut}(H)$ are precisely those of order 3, and by Sylow’s Theorem, every order 3 subgroup is conjugate to $\langle \alpha \rangle$. Let $Z_3 = \langle x \rangle$ and define $\varphi : K \rightarrow \mathsf{Aut}(H)$ by $\varphi(x) = \alpha$. Since $\varphi$ is nontrivial, $Z_{49} \rtimes_\varphi Z_3$ is a nonabelian group of order 147. Suppose now that $\psi : K \rightarrow \mathsf{Aut}(H)$ is some other nontrivial group homomorphism such that $H \rtimes_\psi K$ is nonabelian. Since $K \cong Z_3$ is simple, $\mathsf{ker}\ \psi$ is trivial, so that $\psi$ is injective. Thus $\mathsf{im}\ \psi$ is an order 3 subgroup of $\mathsf{Aut}(H)$, which is conjugate to $\langle \alpha \rangle$. Since $K$ is cyclic, using §5.5 #6 we have $H \rtimes_\psi K \cong H \rtimes_\varphi K$. Thus there exists a unique nonabelian group of order 147 whose Sylow 7-subgroup is cyclic; namely, $Z_{49} \rtimes_\varphi Z_3$. 4. Note first that $|GL_2(\mathbb{F}_7)| = 2^5 \cdot 3^2 \cdot 7$. A straightforward calculation reveals that $t_1^2 = \begin{bmatrix} 4 & 0 \\ 0 & 1 \end{bmatrix}$, $t_1^3 = 1$, $t_2^2 = \begin{bmatrix} 1 & 0 \\ 0 & 4 \end{bmatrix}$, and $t_2^3 = 1$. Moreover, we see that $t_1t_2 = t_2t_1 = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$. Letting $Z_3^2 = \langle a \rangle \times \langle b \rangle$, by this previous exercise there exists a unique group homomorphism $\theta : Z_3^2 \rightarrow \langle t_1,t_2 \rangle$ such that $\theta(a) = t_1$ and $\theta(b) = t_2$. We can see that every element of $\langle t_1, t_2 \rangle$ has the form $t_1^it_2^j$ for some $0 \leq i,j < 3$, so that $|\langle t_1, t_2 \rangle| \leq 9$. Moreover, note the following.  $\theta(1) = I$ $\theta(b) = t_2$ $\theta(b^2) = t_2^2$ $\theta(a) = t_1$ $\theta(ab) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ $\theta(ab^2) = \begin{bmatrix} 2 & 0 \\ 0 & 4 \end{bmatrix}$ $\theta(a^2) = t_1^2$ $\theta(a^2b) = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}$ $\theta(a^2b^2) = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix}$ Thus $\mathsf{ker}\ \theta = 1$, so that $\theta$ is injective, and we have $|P| = 9$. In particular, $P \leq GL_2(\mathbb{F}_7)$ is a Sylow 3-subgroup. Note that if $A \leq GL_2(\mathbb{F}_7)$ is a subgroup of order 3, then $A$ is contained in some Sylow 3-subgroup $Q$ and thus is conjugate to a subgroup of $\langle \alpha \rangle$. 5. Note first that $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$; since $Z_3 = \langle x \rangle$ is cyclic, by §5.5 #6, $G_1 \cong G_2$. Now we find presentations for $G_1$, $G_3$, and $G_4$. Note that each group is generated by $\mu = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, $\eta = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$, and $\omega = x$, and that $\mu \eta = \eta \mu$. 1. $G_1$: We have $\omega \mu = (1,x)(\begin{bmatrix} 1 \\ 0 \end{bmatrix},1)$ $= (\varphi_1(x)(\begin{bmatrix} 1 \\ 0 \end{bmatrix}), x)$ $= (t_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix}, x)$ $= (\begin{bmatrix} 2 \\ 0 \end{bmatrix}, x)$ $= (\begin{bmatrix} 2 \\ 0 \end{bmatrix}, 1)(1, x)$ $= \mu^2 \omega$ and $\omega \eta = (1, x)(\begin{bmatrix} 0 \\ 1 \end{bmatrix}, 1)$ $= (\varphi_1(x)(\begin{bmatrix} 0 \\ 1 \end{bmatrix}), x)$ $= (t_1 \begin{bmatrix} 0 \\ 1 \end{bmatrix}, x)$ $= (\begin{bmatrix} 0 \\ 1 \end{bmatrix}, x)$ $= (\begin{bmatrix} 0 \\ 1 \end{bmatrix}, 1)(1,x)$ $= \eta \omega$. Thus $G_1$ has the presentation $\langle \mu, \eta, \omega \ |\ \mu^7 = \eta^7 = \omega^3, \mu\eta = \eta\mu, \omega\eta = \eta\omega, \omega\mu = \mu^2\omega \rangle$. 2. $G_3$: We have $\omega \mu = (\varphi_3(x)(\begin{bmatrix} 1 \\ 0 \end{bmatrix}), x)$ $= (\begin{bmatrix} 2 \\ 0 \end{bmatrix}, x)$ $= \mu^2 \omega$ and $\omega \eta = (\varphi_3(x)(\begin{bmatrix} 0 \\ 1 \end{bmatrix}),x)$ $= (\begin{bmatrix} 0 \\ 2 \end{bmatrix}, x)$ $= \eta^2 \omega$. Thus $G_3$ has the presentation $\langle \mu, \eta, \omega \ |\ \mu^7 = \eta^7 = \omega^3, \mu\eta = \eta\mu, \omega\eta = \eta^2 \omega, \omega\mu = \mu^2\omega \rangle$. 3. $G_4$: By calculations analogous to those above, we have $\omega \mu = \mu^2 \omega$ and $\omega\eta = \eta^4 \omega$. Thus $G_4$ has the presentation $\langle \mu, \eta, \omega \ |\ \mu^7 = \eta^7 = \omega^3, \mu\eta = \eta\mu, \omega\eta = \eta^4 \omega, \omega\mu = \mu^2\omega \rangle$. 6. We begin with some technical lemmas. Lemma 1: Let $G$ be a group and let $a,b \in G$ such that $ab = b^ka$. Then for all integers $m,n \geq 1$, $a^mb^n = b^{nk^m}a^m$. Proof: We proceed by induction on $m$. 1. For the base case, let $m = 1$. We proceed by induction on $n$. 1. For the base case, let $n = 1$. Clearly then $a^1b^1 = ab = b^ka$ $= b^{1 \cdot k^1}a^1$. 2. For the inductive step, suppose the conclusion holds for some $n \geq 1$. Now $a^1b^{n+1} = a^1 b^nb$ $= b^{nk^1}a^1b$ $= b^{nk^1}b^ka$ $= b^{(n+1)k^1}a^1$. Thus the conclusion holds also for $n+1$. By induction we have $ab^n = b^{nk}a$ for all $n$. 2. For the inductive step, suppose the conclusion holds for some $m \geq 1$. We proceed by induction on $n$. 1. For the base case, let $n = 1$. Now $a^{m+1}b^1 = aa^mb^1$ $= ab^{1 \cdot k^m}a^m$ $= b^{1 \cdot k^m \cdot k} a a^m$ $= b^{1 \cdot k^{m+1}}a^{m+1}$. 2. For the inductive step, suppose the conclusion holds for some $n \geq 1$. Now $a^{m+1}b^{n+1} = aa^mb^nb$ $= ab^{nk^m}a^mb$ $= b^{nk^mk^1} a^1 b^{1 \cdot k^m} a^m$ $= b^{nk^{m+1}}b^{k^{m+1}}a^{m+1}$ $= b^{(n+1)k^{m+1}}a^{m+1}$. By induction we have $a^{m+1}b^n = b^{nk^{m+1}}a^{m+1}$ for all $n$, so that the conclusion holds also for $m+1$. Thus by incution, $a^mb^n = b^{nk^m}a^m$ holds for all integers $m,n \geq 1$. $\square$ Lemma 2: Suppose $m,n$ are positive integers such that $m+a \cdot 2^n \equiv a + m \cdot 2^b$ (mod 7) for all positive integers $a$ and $b$. Then $m \equiv 0$ (mod 7) and $n \equiv 0$ (mod 3). Proof: Let $b = 3$ and $0 < a < 7$. Then $m+a \cdot 2^n \equiv a + m$ (mod 7), so that $a \cdot 2^n \equiv a$ (mod 7). Now $a$ is invertible mod 7, so that $2^n \equiv 1$ (mod 7). Since $|2| = 3$ in $\mathbb{Z}/(7)$, we have $n \equiv 0$ (mod 3). Now we have $m + a \equiv a + m \cdot 2^b$ (mod 7), so that $m \equiv m \cdot 2^b$ (mod 7). If $m \not\equiv 0$ mod 7, then $1 \equiv 2^b$ (mod 7) for all integers $b$, which is absurd. Thus $m \equiv 0$ (mod 7). $\square$ Now we move to the main problem: calculating the centers of $G_1$, $G_3$, and $G_4$. 1. Note from the presentation of $G_1$ that every element in this group can be written in the form $\mu^i \eta^j \omega^k$ for some $0 < i,j \leq 7$ and $0 < k \leq 3$. There are 147 such forms, and because we know by other means that $|G_1| = 147$, in fact each element can be written in this form uniquely. Suppose now that $x = \mu^a \eta^b \omega^c \in Z(G_1)$. Then for every element $\mu^i \eta^j \omega^k$, we have that (using Lemma 1) $\mu^a\eta^b\omega^c \mu^i\eta^j\omega^k = \mu^{a+i2^c} \eta^{b+j} \omega^{c+k}$ is equal to $\mu^i\eta^j\omega^k \mu^a\eta^b\omega^c = \mu^{i + a2^k} \eta^{j+b} \omega^{k+c}$. Thus we have that, for some fixed positive integers $a$ and $c$, $a + i2^c \equiv i + a2^k$ (mod 7)$ for all positive integers $i$ and $k$. By Lemma 2, we have $a \equiv 0$ mod 7 and $c \equiv 0$ mod 3. Thus we have $x \in \langle \eta \rangle$. Clearly we also have $\langle \eta \rangle \leq Z(G_1)$, so that in fact $Z(G_1) = \langle \eta \rangle$.
2. Again, we can see that every element of $G_3$ can be written in the form $\mu^i \eta^j \omega^k$ for some $0 < i,j \leq 7$ and $0 < k \leq 3$, and that because we know that $|G_3| = 147$, every element can be written uniquely in this form. Suppose now that $x = \mu^a \eta^b \omega^c \in Z(G_3)$. Then for every element $\mu^i \eta^j \omega^k$, we have that (using Lemma 1) $\mu^a\eta^b\omega^c \mu^i\eta^j\omega^k = \mu^{a+i2^c} \eta^{b+j2^c} \omega^{c+k}$ is equal to $\mu^i\eta^j\omega^k \mu^a\eta^b\omega^c = \mu^{i + a2^k} \eta^{j+b2^k} \omega^{k+c}$. Comparing exponents and using Lemma 2, we have $a \equiv b \equiv 0$ (mod 7) and $c \equiv 0$ (mod 3). Thus $x = 1$, and we have $Z(G_3) = 1$.
3. Again, we can see that every element of $G_4$ can be written in the form $\mu^i \eta^j \omega^k$ for some $0 < i,j \leq 7$ and $0 < k \leq 3$, and that because we know that $|G_4| = 147$, every element can be written uniquely in this form. Suppose now that $x = \mu^a \eta^b \omega^c \in Z(G_4)$. Then for every element $\mu^i \eta^j \omega^k$, we have that (using Lemma 1) $\mu^a\eta^b\omega^c \mu^i\eta^j\omega^k = \mu^{a+i2^c} \eta^{b+j4^c} \omega^{c+k}$ is equal to $\mu^i\eta^j\omega^k \mu^a\eta^b\omega^c = \mu^{i + a2^k} \eta^{j+b4^k} \omega^{k+c}$. Comparing the exponents of $\mu$, we have by Lemma 2 that $a \equiv 0$ mod 7 and $c \equiv 0$ mod 3. Comparing the exponents of $\eta$, we have $b + j \equiv j + b4^k$ (mod 7), so that $b \equiv b4^k$ (mod 7). If $b \not\equiv 0$ mod 7, then $1 \equiv 4^k$ mod 7 for all $0 < k \leq 3$, which is absurd. Thus $b \equiv 0$ mod 7. Hence we have $x = 1$, and so $Z(G_4) = 1$.

In particular, we see that $G_1 \not\cong G_3$ and $G_1 \not\cong G_4$ since $Z(G_3)$ and $Z(G_4)$ are trivial while $Z(G_1)$ is not.

7. Recall that the Sylow 7-subgroups of $G_3$ and $G_4$ are unique, so that every subgroup of order 7 in each is contained in $H = Z_7^2$. By Example 3 in §5.1, there are 8 such subgroups as follows: $\langle \mu \rangle$, $\langle \mu\eta \rangle$, $\langle \mu\eta^2 \rangle$, $\langle \mu\eta^3 \rangle$, $\langle \mu\eta^4 \rangle$, $\langle \mu\eta^5 \rangle$, $\langle \mu\eta^6 \rangle$, and $\langle \eta \rangle$. Using §3.1 #29, to show that $\langle z \rangle \leq G_i$ is normal it suffices to show that $\omega z \omega^{-1} \in \langle z \rangle$. To that end, note the following. $\omega \eta \omega^{-1} = \eta^2 \in \langle \eta \rangle$ and $\omega \mu \eta^t \omega^{-1} = \mu^2 \omega \eta^t \omega^{-1}$ $= \mu^2 \eta^{2t}$ $= (\mu \eta^t)^2$ $\in \langle \mu \eta^t \rangle$. Thus every subgroup of $G_3$ of order 7 is normal there. On the other hand, in $G_4$, we have $\omega \mu \eta \omega^{-1} = \mu^2 \omega \eta \omega^{-1}$ $= \mu^2 \eta^4$ $= (\mu \eta^2)^2$ $\in \langle \mu \eta^2 \rangle$. Since (by Lagrange) $\langle \mu\eta \rangle \cap \langle \mu\eta^2 \rangle = 1$, $\omega\mu\eta\omega^{-1} \notin \langle \mu\eta \rangle$. Thus $\langle \mu\eta \rangle$ is an order 7 subgroup which is not normal in $G_4$. Thus $G_3 \not\cong G_4$.
8. Let $G$ be a group of order $147 = 3 \cdot 7^2$. If $G$ is abelian, then $G$ is isomorphic to one of $Z_{147}$ and $Z_{21} \times Z_7$.

If $G$ is nonabelian, then by part (2) above, $G$ has a normal Sylow 7-subgroup $H$. Let $K \cong Z_3$ be any Sylow 3-subgroup of $H$. By Lagrange, $H \cap K = 1$, so that $G = HK$. By the recognition theorem for semidirect products, $G \cong H \rtimes_\varphi K$, where $K = Z_3$, $|H| = 49$, and $\varphi : K \rightarrow \mathsf{Aut}(H)$. Thus classifying the nonabelian groups of order 147 is evidently equivalent to determining the nonisomorphic groups constructed in this manner.

By Part (3) above, there is an essentially unique nonabelian group of the form $Z_{49} \rtimes_\varphi Z_3$.

Suppose now that $H = Z_7^2$. Since $G$ is nonabelian, $\varphi$ is nontrivial. Since $K$ is simple, we have $\mathsf{ker}\ \varphi = 1$; thus $\mathsf{im}\ \varphi$ is a subgroup of order 3 in $\mathsf{Aut}(H) \cong GL_2(\mathbb{F}_7)$. By part (4) above, $\mathsf{im}\ \varphi$ is conjugate to one of the four subgroups of the Sylow 3-subgroup $P \leq GL_2(\mathbb{F}_7)$ identified above. Since $K$ is cyclic, using this previous exercise, $H \rtimes_\varphi K$ is isomorphic to one of the groups $G_i$ identified above. We showed in Parts (5), (6), and (7) that three of these are distinct.

Thus the groups of order 147 up to isomorphism are as follows. In each case, we let $Z_3 = \langle x \rangle$, $Z_{49} = \langle y \rangle$, and $Z_7^2 = \langle a \rangle \times \langle b \rangle$.

1. $Z_{147}$
2. $Z_{21} \times Z_7$
3. $Z_{49} \rtimes_\psi Z_3$, where $\psi(x)(y) = y^{-4}$.
4. $Z_7^2 \rtimes_{\varphi_1} Z_3$, where $\varphi_1(x)(a) = a^2$ and $\varphi_1(x)(b) = b$
5. $Z_7^2 \rtimes_{\varphi_2} Z_3$, where $\varphi_2(x)(a) = a^2$ and $\varphi_2(x)(b) = b^2$
6. $Z_7^2 \rtimes_{\varphi_3} Z_3$, where $\varphi_3(x)(a) = a^2$ and $\varphi_3(x)(b) = b^4$.