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<j, then for all k<i, 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.