## Monthly Archives: June 2010

### Equivalent characterizations of nilpotent and cyclic groups

If $G$ is finite, prove that (1) $G$ is nilpotent if and only if it has a normal subgroup of each order dividing $|G|$, and (2) is cyclic if and only if it has a unique subgroup of each order dividing $|G|$.

We begin with some lemmas.

Lemma 1: Let $G$ be a group, and let $H_1, \ldots, H_k$ be characteristic subgroups of $G$. Then $H = H_1\cdots H_k$ is characteristic in $G$. Proof: $H$ is a subgroup because each $H_i$ is normal. If $\alpha$ is an automorphism of $G$, then $\alpha[H] = \alpha[H_1] \cdots \alpha[H_k]$ $= H_1 \cdots H_k = H$. $\square$

Lemma 2: Let $G$ be a finite group. If $P \leq G$ is a normal Sylow subgroup, then $P$ is characteristic. Proof: $P$ is the unique subgroup of order $|P|$. $\square$

Now to the main result; first we prove (1).

$(\Rightarrow)$ We proceed by induction on the breadth (number of prime divisors with multiplicity) of $G$.

For the base case, if $G$ is a finite nilpotent group of breadth 1, then $G \cong Z_p$ is an abelian simple group. Thus $G$ has a normal subgroup of order 1 and $p$, the only divisors of $|G|$.

For the inductive step, suppose that every finite nilpotent group of breadth $k \geq 1$ has a normal subgroup of each order dividing $|G|$, and let $G$ be a finite nilpotent group of breadth $k+1$. Now $Z(G)$ is nontrivial, so there exists an element $x \in Z(G)$ of prime order, and $\langle x \rangle G$ is normal. We showed in a lemma to the previous exercise that $G/\langle x \rangle$ is nilpotent, and has breadth $k$. By the inductive hypothesis, $G/\langle x \rangle$ has a normal subgroup of each order $d$ dividing $|G/\langle x \rangle|$. By the Lattice Isomorphism Theorem, a subgroup $\overline{H} \leq G/\langle x \rangle$ is normal if and only if $\overline{H} = H/\langle x \rangle$ for some normal subgroup $H \leq G$. Thus we see that $G$ has a normal subgroup of order $d$ for all $d$ dividing $|G|$ such that $p|d$. Now let $d$ divide $|G|$, with $p$ not dividing $d$, and let $H \leq G$ be a normal subgroup of order $pd$. Since $H \leq G$, $H$ is a finite nilpotent group. By Theorem 3, $H$ is the direct product of its Sylow subgroups. Write $H = P \times Q$, where $P$ is the Sylow $p$ subgroup of $H$ and $Q$ the direct product of the remaining Sylow subgroups. By Lemma 2, each factor of $Q$ is characteristic in $H$, so that $Q$ is characteristic in $H$. Note that $|Q| = d$, thus $Q \leq G$ is normal and has order $d$.

$(\Leftarrow)$ Suppose $G$ has a normal subgroup of each order $d$ dividing $|G|$. In particular, $G$ has a normal (thus unique) Sylow $p$-subgroup for each prime $p$. By Theorem 6.3, $G$ is nilpotent.

Now we prove (2).

The $(\Rightarrow)$ direction was proved in Theorem 2.7.

$(\Leftarrow)$ Suppose $G$ is finite and has a unique subgroup of order $d$ for each $d$ properly dividing $|G|$. In particular, all the Sylow subgroups of $G$ are unique, hence normal; say there are $k$ of these. Suppose some Sylow $p$-subgroup $P$ of $G$ is not cyclic; then by this previous exercise, $P$ has a subgroup congruent to $Z_p \times Z_p$, and thus $G$ has distinct subgroups of order $p$, a contradiction. Thus each Sylow subgroup of $G$ is cyclic; say $P_i = \langle x_i \rangle$. Then $|x_1 \cdots x_k| = |G|$, so that $G = \langle x_1 \cdots x_k \rangle$ is cyclic.

### Basic properties of finite nilpotent groups

Let $G$ be a finite nilpotent group.

1. If $H$ is a nontrivial normal subgroup of $G$, then $H$ intersects the center of $G$ nontrivially. In particular, every normal subgroup of prime order is central.
2. If $H < G$ is a proper subgroup then $H$ is properly contained in $N_G(H)$.

First we prove some lemmas.

Lemma 1: Let $G$ be a group, with $A_1,A_2,B_1,B_2 \subseteq G$. If $A_1 \subseteq A_2$ and $B_1 \subseteq B_2$, then $[A_1,B_1] \leq [A_2,B_2]$. Proof: $[A_1,B_1]$ is generated by $[a,b]$ such that $a \in A_1$ and $b \in B_1$. Each of these generators is in $[A_2,B_2]$. $\square$

Lemma 2: Let $G$ be a group. If $H \leq G$, then $H^k \leq G^k$ for all $k$, where $G^k$ denotes the $k$-th term in the lower central series of $G$. Proof: We proceed by induction on $k$. For the base case, if $k=0$ then $H^0 \leq G^0$. For the inductive step, suppose $H^k \leq G^k$ for some $k \geq 0$. Then by Lemma 1, $H^{k+1} = [H,H^k] \leq [G,G^k] = G^{k+1}$. $\square$

Lemma 3: If $G$ is a nilpotent group of nilpotence class $k$ and $H \leq G$, then $H$ is nilpotent with nilpotence class at most $k$. Proof: By Theorem 8 in the text, $G^k = 1$. By Lemma 2, $H^k = 1$, so that $H$ is nilpotent and of nilpotence class at most $k$. $\square$

Lemma 4: Let $\varphi : G \rightarrow H$ be a group homomorphism and let $S \subseteq G$. Then $\varphi[\langle S \rangle] = \langle \varphi[S] \rangle$. Proof: We have $x \in \varphi[\langle S \rangle]$ if and only if $x = \varphi(s_1^{a_1} \cdots s_k^{a_k})$ for some $s_i \in S$ if and only if $x = \varphi(s_1)^{a_1} \cdots \varphi(s_k)^{a_k}$ for some $s_i \in S$ if and only if $x \in \langle \varphi[S] \rangle$. $\square$

Lemma 5: Let $\varphi : G \rightarrow H$ be a surjective group homomorphism. Then $H^k \leq \varphi[G^k]$ for all $k$, where $G^k$ denotes the $k$-th term in the lower central series of $G$. Proof: We proceed by induction on $k$. For the base case, if $k = 0$ then $H^0 = H = \varphi[G] = \varphi[G^0]$. For the inductive step, suppose $H^k \leq \varphi[G^k]$. Then we have $H^{k+1} = [H,H^k] \leq [\varphi[G],\varphi[G^k]]$ $\leq \varphi[[G,G^k]]$ $\leq \varphi[G^{k+1}]$. $\square$

Lemma 6: Let $G$ be a nilpotent group of nilpotence class $k$. If $H \leq G$ is normal, then $G/H$ is nilpotent of nilpotence class at most $k$. Proof: By Lemma 5, we have $(G/H)^k \leq \varphi[G^k] = 1$. $\square$

Now to the main results; first we show part 2. [With hints from MathReference.]

(2) Let $G$ be a finite nilpotent group such that $Z(G) \neq 1$. If $H \leq G$ is a nontrivial normal subgroup of $G$, then $H \cap Z(G)$ is nontrivial. In particular, every normal subgroup of order $p$ is contained in the center.

We proceed by induction on the breadth of $G$. (Recall: the breadth of a finite group is the number of prime factors dividing its order, including multiplicity.)

For the base case, if $G$ has breadth 1, then $G \cong Z_p$ for some prime $p$ and thus $G$ is simple and abelian. Thus $H = G$, and $H \cap Z(G) \neq 1$.

For the inductive step, suppose that the conclusion holds for every finite nilpotent group of breadth at most $k \geq 1$, and let $G$ be a finite nilpotent group of breadth $k+1$. Let $H \leq G$ be normal and suppose $H \cap Z(G) = 1$. Now $H \cdot Z(G) \cong H \times Z(G) \leq G$ is an internal direct product. Moreover, since $H$ and $Z(G)$ are normal in $G$, $H \cdot Z(G) \leq G$ is normal. Now by the Lattice Isomorphism Theorem, $HZ(G)/Z(G) \leq G/Z(G)$ is normal. By Lemma 6, $G/Z(G)$ is nilpotent, has breadth at most $k$, and $HZ(G)/Z(G) \leq G/Z(G)$ is normal. By the inductive hypothesis, there exists an element $xZ(G) \in Z(G/Z(G))$ such that $xZ(G) \in HZ(G)/Z(G)$. We can write $x = hz$, where $h \in H$ and $z \in Z(G)$, so that in fact $hZ(G) \in Z(G/Z(G))$. Now let $g \in G$ be arbitrary; since $H$ is normal in $G$, $ghg^{-1} \in H$. Now $(ghg^{-1})Z(G) = (gZ(G))(hZ(G))(g^{-1}Z(G))$ $= hZ(G)$ since $hZ(G)$ is in the center of $G/Z(G)$, so that $ghg^{-1}Z(G) = hZ(G)$. Write $ghg^{-1} = hk$, where $k \in Z(G)$. Now $ghg^{-1}h^{-1} = k \in Z(G)$, and since $H \cap Z(G) = 1$, we have $ghg^{-1}h^{-1} = 1$, thus $gh = hg$. So $h \in H$ is in the center of $G$ and $h \neq 1$, a contradiction. Thus $H \cap Z(G)$ is nontrivial. $\square$

Next, we prove some more lemmas.

Lemma 7: Let $G$ be a group, $K, H_1, H_2 \leq G$ be subgroups with $K \leq G$ normal, $K \leq H_1$, and $K \leq H_2$. If $H_1/K = H_2/K$, then $H_1 = H_2$. Proof: Let $x \in H_1$. Then $xK \in H_1/K = H_2/K$, so that $xK = yK$. In particular, $x \in yK \leq H_2$. The other direction is similar. $\square$

Lemma 8: Let $G$ be a group, $H \leq G$ a normal subgroup, and $K$ a subgroup with $H \leq K \leq G$. Then $N_{G/H}(K/H) = N_G(K)/H$. Proof: Note that $xH \in N_{G/H}(K/H)$ if and only if $(xH)(K/H)(x^{-1}H) = K/H$, if and only if $(xKx^{-1})/H = K/H$, if and only if (by Lemma 7) $xKx^{-1} = K$, if and only if $x \in N_G(K)$, if and only if $xH \in N_G(K)/H$. $\square$

Now we show part 4.

(4) Let $G$ be a finite nilpotent group. If $H < G$ is a proper subgroup, then $H < N_G(H)$ is proper.

Proof: We proceed again by induction on the breadth of $G$.

For the base case, if $G$ has breadth 1, then $G \cong Z_p$ is abelian. Then if $H \leq G$ is proper, $H \leq N_G(H) = G$ is proper.

For the inductive step, suppose the conclusion holds for all finite nilpotent groups of breadth at most $k$. Let $G$ be a finite nilpotent group of breadth $k+1$ and let $H \leq G$ be a proper subgroup. Now $Z(G) \leq N_G(H)$. If $Z(G) \not\leq H$, then $H \leq \langle H, Z(G) \rangle$ is proper, so that $H \leq N_G(H)$ is proper. If $Z(G) \leq H$, then $H/Z(G) \leq G/Z(G)$ is proper. Moreover, $G/Z(G)$ is a finite nilpotent group of breadth at most $k$, so that $H/Z(G) \leq N_{G/Z(G)}(H/Z(G))$ is proper. By Lemma 8, $H/Z(G) \leq N_G(H)/Z(G)$ is proper, so that $H \leq N_G(H)$ is proper. $\square$

### The higher centers in a group are characteristic

Prove that $Z_k(G)$ is characteristic in $G$ for all $i$.

We begin with a lemma.

Lemma 1: Let $G$ be a group, let $K \leq G$ be characteristic, and let $\varphi$ be an automorphism of $G$. Then the mapping $\psi : G/K \rightarrow G/K$ given by $\psi(aK) = \varphi(a)K$ is an automorphism of $G/K$. Proof: (Well-defined) Suppose $aK = bK$; then $ab^{-1} \in K$. Since $K$ is characteristic, $\varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1} \in K$, so that $\varphi(a)K = \varphi(b)K$. Thus $\psi(aK) = \psi(bK)$. (Homomorphism) We have $\psi((ab)K) = \varphi(ab)K$ $= (\varphi(a)K)(\varphi(b)K)$ $= \psi(aK)(\psi(bK)$. (Injective) Suppose $\psi(aK) = \psi(bK)$. Then $\varphi(ab^{-1}) \in K$, so that (since $K$ is characteristic) $aK = bK$. (Surjective) Let $aK \in G/K$. Since $\varphi$ is surjective, $aK = \varphi(b)K = \psi(bK)$ for some $b \in G$. $\square$

Lemma 2: Let $G$ be a group. Suppose $K \leq H \leq G$ such that $K \leq G$ is characteristic and $H/K \leq G/K$ is characteristic. Then $H \leq G$ is characteristic. Proof: Let $\varphi \in \mathsf{Aut}(G)$. By Lemma 1, the mapping $\psi : G/K \rightarrow G/K$ given by $\psi(aK) = \varphi(a)K$ is an automorphism of $G/K$. Since $H/K$ is characteristic, we have $\psi[H/K] = H/K$; that is, if $h \in H$, then $\psi(hK) = \varphi(h)K = xK$ for some $x \in H$. Thus $\varphi(h)x^{-1} \in K$, and we have $\varphi(h) \in H$. Thus $\varphi[H] \leq H$. Now suppose $h \in H$. Since $H/K$ is characteristic in $G/K$, we have $hK = \psi(xK) = \varphi(x)K$ for some $x \in H$. Then $\varphi(x^{-1})h = k \in K$. Since $K$ is characteristic in $G$, we have $k = \varphi(y)$ for some $y \in K$. Thus $h = \varphi(xy)$, where $xy \in HK = H$. Thus we have $H \leq \varphi[H]$. Hence $H \leq G$ is characteristic. $\square$

We proceed by induction on $k$.

For the base case, if $k=0$, then $Z_k(G) = 1$. Clearly then $Z_k(G)$ is characteristic in $G$.

For the inductive step, let $k \geq 0$ and suppose that $Z_k(G) \leq G$ is characteristic. Consider $Z_{k+1}(G) \leq G$. By definition, $Z_{k+1}(G)$ is the subgroup of $G$ containing $Z_k(G)$ such that $Z_{k+1}(G)/Z_k(G) = Z(G/Z_k(G))$. Now by definition, $Z_{k+1}(G)/Z_k(G)$ is characteristic in $G/Z_k(G)$. By Lemma 2, $Z_{k+1}$ is characteristic in $G$.

By induction, $Z_k(G)$ is characteristic in $G$ for all $k$.

### 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 $H(\mathbb{F}_p) \cong (Z_p \times Z_p) \rtimes Z_p$.

### Characterize the natural numbers which are only the orders of abelian groups

Let $n > 1$ be an integer. Prove the following classification: every group of order $n$ is abelian if and only if $n = p_1^{a_1} \cdots p_k^{a_k}$, $p_i$ are distinct primes, $a_i \in \{1,2\}$, and $p_i$ does not divide $p_j^{a_j}-1$ for all $i$ and $j$.

$(\Rightarrow)$ Suppose $n$ has the properties above; we proceed by induction on $k$.

For the base case, let $k = 1$ and let $G$ be a group of order $n = p_1^{a_1}$. Then $G$ is either cyclic or elementary abelian, hence $G$ is abelian.

For the inductive step, suppose that every group of width at most $k$ which satisfies the given conditions for $|G|$ is abelian. Let $G$ be a group of width $k+1$ such that $|G|$ satisfies the given properties. Now every proper subgroup of $G$ has width at most $k$, so that by the induction hypothesis, every proper subgroup of $G$ is abelian. By this previous exercise, $G$ is solvable.

By this previous exercise, there is a (maximal) subgroup $H \leq G$ such that $H \leq G$ is normal and $G/H$ has prime order; say $G/H \cong Z_p$. Let $P$ be a Sylow $p$-subgroup of $G$. Now by FTFGAG, $H$ is the direct product of its Sylow subgroups, and the Sylow subgroups of $H$ are normal in $H$. In particular, $H$ is generated by elements of prime and prime squared order. Let $x \in H$ have prime or prime squared order and let $y \in P$. If $\langle xy \rangle = G$, then $G$ is cyclic, hence abelian. If $\langle xy \rangle$ is a proper subgroup of $G$, then $xy = yx$, so that $\langle x \rangle \leq C_G(y)$. In either case, we have $\langle x \rangle \leq C_G(P)$. Thus $H \leq C_G(P)$, and we have $HP = G \leq C_G(P)$. Thus $P \leq Z(G)$. Similarly, $HP = G \leq C_G(H)$, so that $H \leq Z(G)$. Thus $HP = G \leq Z(G)$, and $G$ is abelian.

$(\Leftarrow)$ Suppose every group of order $n = p_1^{a_1} \cdots p_k^{a_k}$ is abelian. If (without loss of generality) $a_1 \geq 3$, let $Q$ be a nonabelian group of order $p_1^{a_1}$. (We may take $Q$ to be the direct product of a nonabelian group of order $p^3$ and a cyclic group.) Then $Q \times Z_{p_2^{a_2} \cdots p_k^{a_k}}$ is nonabelian of order $n$. Thus $a_i \in \{1,2\}$ for all $i$. Now suppose (again without loss of generality) that $p_1$ divides $p_2^{a_2} - 1$. If $a_2 = 1$, then we may construct a nonabelian group $Q$ of order $p_1p_2$. Then $Q \times Z_{p_3^{a_3} \cdots p_k^{a_k}}$ is a nonabelian group of order $n$. If $a_2 = 2$, we may construct a nonabelian group of order $pq^2$, following the patterns here or here (depending on whether $p_1$ does not or does divide $p_2-1$.) By taking the direct product with a cyclic group, there exists a nonabelian group of order $n$.

Thus, if every group of order $n$ is abelian, then $n$ has the properties enumerated in the problem statement.

### Basic properties of wreath products

Let $K$ and $L$ be groups, let $n$ be a positive integer, let $\rho : K \rightarrow S_n$ be a homomorphism, and let $H = L^n$. In this previous exercise, we defined an injective homomorphism $\psi : S_n \rightarrow \mathsf{Aut}(H)$ by permuting the factors. Now $\psi \circ \rho$ is a homomorphism $K \rightarrow \mathsf{Aut}(H)$. The wreath product of $L$ by $K$ via $\rho$ is defined $L \wr_\rho K = H \rtimes_{\psi \circ \rho} K$. Note that the wreath product depends on the choice of permutation representation; if none is given, $\rho$ is assumed to be the left regular representation.

1. Assume $L$ and $K$ are finite and $\rho$ is the left regular representation of $K$. Find $|L \wr_\rho K|$ in terms of $|K|$ and $|L|$.
2. Let $p$ be a prime, let $K = L = Z_p$, and let $\rho$ be the left regular representation of $K$. Prove that $Z_p \wr_\rho Z_p$ is a nonabelian group of order $p^{p+1}$ and is isomorphic to a Sylow $p$-subgroup of $S_{p^2}$. [The $p$ copies of $Z_p$ whose direct product makes up $H$ may be represented by $p$ disjoint $p$-cycles; these are cyclically permuted by $K$.]

1. We have $L \wr_\rho K = L^{|K|} \rtimes K$. Thus $|L \wr_\rho K| = |L|^{|K|} \cdot |K|$.
2. Since the left regular representation is nontrivial, and $\psi$ is nontrivial, $Z_p \wr_\rho Z_p$ is a nonabelian group. By part (1), this group has order $p^p \cdot p = p^{p+1}$.

Now $H = Z_p^p$ is a finite abelian group. Let $\sigma_i$, $1 \leq i \leq p$, be $p$ disjoint $p$-cycles in $S_{p^2}$. By this previous exercise, there is a unique group homomorphism $\varphi : H \rightarrow S_{p^2}$ such that $\varphi(e_i) = \sigma_i$, where $e_i$ is the $i$th “standard basis” vector. Moreover, $(\psi \circ \rho)(k)(e_i) = e_{i+k}$. By our solution to this previous exercise, there exists a permutation $\tau \in S_{p^2}$ such that $\tau \sigma_i \tau^{-1} = \sigma_{i+1}$, where indices are taken mod $p$. Define $\theta : Z_p \wr_\rho Z_p \rightarrow S_{p^2}$ by $\theta(h,k) = \varphi(h)\tau^{-k}$.

I claim that $\theta$ is a homomorphism. To see this, write $h_1 = \prod e_i^{a_i}$ and $h_2 = \prod e_i^{b_i}$. Then $\theta((h_1,k_1)(h_2,k_2)) = \theta(((\prod e_i^{a_i})(\psi \circ \rho)(k_1)(\prod e_i^{b_i}), k_1k_2))$ $= \theta(((\prod e_i^{a_i})(\prod e_{i+k_1}^{b_i}), k_1k_2))$ $= \theta((\prod e_i^{a_i + b_i - k_1}), k_1 k_2)$ $= \prod \sigma_i^{a_i + b_i - k_1} \tau^{-k_1-k_2}$ $= (\prod \sigma_i^{a_i})\tau^{-k_1} (\prod \sigma_i^{b_i}) \tau^{-k_2}$ $= \theta(h_1,k_1) \theta(h_2,k_2)$.

Moreover, I claim that $\theta$ is injective; in this previous exercise we saw that every element of $\mathsf{im}\ \theta$ can be written uniquely as a product $(\prod \sigma_i^{a_i}) \tau^k$. Thus $\theta$ is injective. Counting elements, we see that $\mathsf{im}\ \theta$ is a Sylow $p$-subgroup of $S_{p^2}$.

### 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 $\varphi(\alpha,\beta)(x) = \alpha x \beta^{-1}$.

### For odd primes p, a p-group whose every subgroup is normal is abelian

Let $p$ be an odd prime and let $P$ be a $p$-group. Prove that if every subgroup of $P$ is normal then $P$ is abelian. (Note that $Q_8$ is a nonabelian 2-group with this property, so the result fails for $p=2$.)

[Disclaimer: I looked at Alfonso Gracia-Saz’s notes when solving this problem.]

We proceed by induction on $k$ such that $|P| = p^k$.

For the base case, if $k = 1$, then $P$ is cyclic and thus abelian.

For the inductive step, suppose the conclusion holds for all $P$ such that $|P| = p^k$, where $k \geq 1$. Let $P$ be a group of order $p^{k+1}$. If $P$ is cyclic, then $P$ is abelian. If $P$ is not cyclic, then by the previous exercise there exists a normal subgroup $U \leq P$ with $U \cong Z_p \times Z_p$. Choose $x,y \in U$ such that $U = \langle x \rangle \times \langle y \rangle$. By the hypothesis, $\langle x \rangle, \langle y \rangle \leq P$ are normal. Now $P/\langle x \rangle$ and $P/\langle y \rangle$ are groups of order $p^k$, and by the Lattice Isomorphism Theorem and the induction hypothesis, every subgroup of $P/\langle x \rangle$ and $P/\langle y \rangle$ is normal; thus $P/\langle x \rangle$ and $P/\langle y \rangle$ are abelian. By this previous exercise, $P/(\langle x \rangle \cap \langle y \rangle) \cong P$ is abelian.

### With p an odd prime, every noncyclic p-group contains a normal direct product of two copies of Cyc(p)

Let $p$ be an odd prime. Prove that if $P$ is a noncyclic $p$-group then $P$ contains a normal subgroup $U$ with $U \cong Z_p \times Z_p$. Deduce that for odd primes $p$ a $p$-group that contains a unique subgroup of order $p$ is cyclic. (Note: for $p = 2$, it is a theorem that the generalized quaternion groups $Q_{2^n}$ are the only noncyclic 2-groups which contain a unique subgroup of order 2.)

Let $p$ be an odd prime and let $P$ be a $p$-group.

We will proceed by induction on the breadth of $P$; that is, on the number $k$ such that $|P| = p^k$.

For the base case, let $k = 2$ and let $P$ be a noncyclic group of order $p^2$. Then $P \cong Z_p \times Z_p$ by Corollary 4.9. Then $U = P$ is a normal subgroup of $P$ isomorphic to $Z_p \times Z_p$.

For the inductive step, suppose that for some $k \geq 2$, every noncyclic group of order $p^k$ contains a normal subgroup isomorphic to $Z_p \times Z_p$. Let $P$ be a noncyclic group of order $p^{k+1}$. Since $P$ is a $p$-group, $Z(P)$ is nontrivial. By Cauchy’s Theorem, there exists an element $x \in Z(P)$ of order $p$, and $Z = \langle x \rangle$ is normal in $P$. Now $P/Z$ is a $p$-group of order $p^k$.

Suppose $P/Z$ is cyclic. By the Third Isomorphism Theorem, we have $(P/Z)/(Z(P)/Z) \cong P/Z(P)$. Since $P/Z$ is cyclic, every quotient of $P/Z$ is cyclic, so that $P/Z(P)$ is cyclic. By this previous exercise, $P$ is abelian. Now $P$ is a finite abelian group of rank at least 2, since $P$ is not cyclic. Let $\varphi : P \rightarrow P$ be the $p$-power map. By this previous exercise, $\mathsf{ker}\ \varphi$ is the elementary abelian $p$-group whose rank is the same as that of $P$; that is, $\mathsf{ker}\ \varphi \cong Z_p^t$ where $t \geq 2$. Clearly $\mathsf{ker}\ \varphi$ contains a subgroup $U$ isomorphic to $Z_p \times Z_p$, and since $U \leq P$ and $P$ is abelian, $U$ is normal in $P$.

Suppose $P/Z$ is not cyclic. By the induction hypothesis, there is a normal subgroup $\overline{H} \leq P/Z$ such that $\overline{H} \cong Z_p \times Z_p$. By the Fourth (Lattice) Isomorphism Theorem, there is a normal subgroup $H \leq P$ such that $\overline{H} = H/Z$. In particular, $H$ is a noncyclic group of order $p^3$.

Now let $\varphi : H \rightarrow H$ be the $p$-power map. By this previous exercise, $\varphi$ is a homomorphism and in fact $\mathsf{im}\ \varphi \leq Z(H)$. Note that $H$ is not cyclic since $H/Z$ is not cyclic. Thus, also by this exercise, $\mathsf{ker}\ \varphi$ has order $p^2$ or $p^3$. Moreover, note that if $x \in \mathsf{ker}\ \varphi$ and $\alpha \in \mathsf{Aut}(H)$, then $x$ has order $p$ and thus $\alpha(x)$ has order $p$. In particular, $\alpha(x) \in \mathsf{ker}\ \varphi$. So $\alpha[\mathsf{ker}\ \varphi] \subseteq \mathsf{ker}\ \varphi$, and since $H$ is finite, in fact $\alpha[\mathsf{ker}\ \varphi] = \mathsf{ker}\ \varphi$. Thus $\mathsf{ker}\ \varphi$ is characteristic in $H$, hence normal in $P$.

If $|\mathsf{ker}\ \varphi| = p^2$, then $\mathsf{ker}\ \varphi$ is (isomorphic to) the elementary abelian group of order $p^2$; thus $U = \mathsf{ker}\ \varphi \cong Z_p \times Z_p$ is normal in $P$.

[With a hint from Alfonso Gracia-Saz’s notes] Suppose $|\mathsf{ker}\ \varphi| = p^3$; then in fact $H = \mathsf{ker}\ \varphi$. Thus $H$ is the elementary abelian group of order $p^3$, and $H/Z \cong Z_p \times Z_p$. Let $\mathcal{H}$ denote the set of all subgroups $K/Z \leq H/Z$ of order $p$. Note that $P/Z$ acts on $\mathcal{H}$ by conjugation, and that $|\mathcal{H}| = p+1$. Now given $K/Z \in \mathcal{H}$, by the Orbit-Stabilizer Theorem, we have $|P/Z \cdot K| = [P/Z : \mathsf{stab}_{P/Z}(K/Z)]$; since $P/Z$ is a $p$-group, each orbit thus has order a power of $p$. Since $|\mathcal{H}| = p+1$, there is thus one orbit of order $p$ and one orbit of order 1. Let $N/Z$ be in the orbit of order 1. That is, $N/Z$ is normal in $P/Z$, and by the Fourth Isomorphism Theorem, $N \leq P$ is normal. Finally, $|N| = p^2$, and $N \leq H$ is elementary abelian. Thus $U = N \cong Z_p \times Z_p$ has the desired properties.

Finally, suppose $P$ is a $p$ group which contains a unique subgroup of order $p$. If $P$ is not cyclic, then by the above argument $P$ has a subgroup isomorphic to $Z_p \times Z_p$, yielding a contradiction because this subgroup contains many subgroups isomorphic to $Z_p$. Thus $P$ is cyclic.

### Some basic properties of holomorphs

Let $H$ be a group of order $n$, let $K = \mathsf{Aut}(H)$, and form $G = \mathsf{Hol}(H) = H \rtimes_1 K$. Let $G$ act by left multiplication on the left cosets of $K$ in $G$ and let $\pi : G \rightarrow S_{G/K}$ be the induced permutation representation.

1. Prove that the elements of $H$ are coset representatives for the left cosets of $K$ in $G$ and that with this choice of representatives, $\pi|_H$ is the left regular representation of $H$.
2. Prove that $\mathsf{im}\ \pi$ is the normalizer in $S_{G/K}$ of $\pi[H]$. Deduce that under the left regular representation of a finite group $H$ of order $n$, the noramlizer in $S_n$ of the image of $H$ is isomorphic to $\mathsf{Hol}(H)$. [Hint: Show that $|G| = |N_{S_n}(\pi[H])|]$ using §5.5 #1 and §5.5 #2.
3. Deduce that the normalizer of the group generated by an $n$-cycle in $S_n$ is isomorphic to $\mathsf{Hol}(Z_n)$ and has order $n \varphi(n)$, where $\varphi$ denotes the Euler totient function.

1. Recall that $(h,k) = (h,1)(1,k)$. It suffices to show that if $h_1,h_2 \in H$ such that $h_1K = h_2K$, then $h_1 = h_2$. To that end, if $(h_1,1) = (h_2,k)$, then $h_1 = h_2$. There are $n$ left cosets of $K$, and $H$ maps to these injectively. Thus $H$ is a set of coset representatives of $G/K$. Define $\theta : S_{G/K} \rightarrow S_H$ by $\theta(hK) = h$, and let $\rho : H \rightarrow S_H$ be the left regular representation of $H$. Then $\rho(h)(a) = ha = (ha)K$ $h(aK)$ $= \pi(h)(aK)$ $= (\pi(h) \circ \theta)(a)$. Hence $\rho(h) = \pi(h) \circ \theta$; thus $\pi[H]$ is the left regular representation of $H$ up to an isomorphism.
2. First, let $\pi(g) \in \pi[G]$. Now $\pi(g)\pi[H]\pi(g)^{-1} = \pi[gHg^{-1}]$ $= \pi[H]$. Thus $\pi(g) \in N_{S_{G/K}}(\pi[H])$.

Now we prove a lemma.

Lemma [Adapted from Wikipedia]: If $\sigma \in N_{S_{G/K}}(\pi[H])$ and $\sigma(1) = 1$, then $\sigma \in \mathsf{Aut}(H)$. Proof: Since $\sigma$ normalizes $\pi[H]$, for every $h \in H$ there exists $k \in H$ such that $\sigma \circ \pi(h) = \pi(k) \circ \sigma$. Applying this mapping to 1, we see that $\sigma(h) = k$. Thus, if $\sigma(1) = 1$, then for all $h \in H$ we have $\sigma \circ \pi(h) = \pi(\sigma(h)) \circ \sigma$. Now for all $h,k \in H$, we have $\sigma \circ \pi(h) \circ \pi(k) = \pi(\sigma(h)) \circ \sigma \circ \pi(k)$ $= \pi(\sigma(h)) \circ \pi(\sigma(k)) \circ \sigma$ $= \pi(\sigma(h) \sigma(k)) \circ \sigma$, which is equal (since $\pi$ is a homomorphism) to $\sigma \circ \pi(hk) = \pi(\sigma(h,k)) \circ \sigma$. Since $\sigma$ is invertible and $\pi$ is injective, we have $\sigma(hk) = \sigma(h) \sigma(k)$; hence $\sigma$ is a homomorphism $H \rightarrow H$. Since $\sigma$ is also bijective, $\sigma \in \mathsf{Aut}(H)$. $\square$

$(\supseteq)$ Now let $\sigma \in N_{S_{G/K}}(\pi[H])$. By the argument above, $\pi(\sigma(1)^{-1})$ also normalizes $\pi[H]$. Since the normalizer is a subgroup, we have $\pi(\sigma^{-1}(1)) \circ \sigma \in N_{S_{G/K}}(\pi[H])$. Moreover, $(\pi(\sigma^{-1}(1)) \circ \sigma)(1) = \pi(\sigma(1)^{-1})(\sigma(1)) = 1$; by the lemma, $\beta = \pi(\sigma^{-1}(1)) \circ \sigma$ is an automorphism of $H$. Finally, note that $\pi(\beta)(h) = \beta(h)$, so that $\beta = \pi(\beta)$. Thus we have $\sigma = \pi(\sigma(1)^{-1})^{-1} \circ \beta$ $= \pi(\sigma(1)) \circ \pi(\beta)$ $= \pi(\sigma(1),\beta)$. Hence $\sigma \in \pi[G]$.

Since $\pi$ is injective, we conclude that $\mathsf{Hol}(H) \cong N_{S_H}(\mathsf{im}\ \rho)$.

3. Let $\tau$ be an $n$-cycle in $S_n$. Up to a renaming, $\langle \tau \rangle$ is the image of $Z_n$ in $S_{Z_n}$ via the permutation representation $\pi : \mathsf{Hol}(Z_n) \rightarrow S_n$ used above. Now $N_{S_n}(\langle \tau \rangle) \cong \mathsf{Hol}(Z_n)$, and $\mathsf{Hol}(Z_n) = Z_n \rtimes \mathsf{Aut}(Z_n)$. We know that $|\mathsf{Aut}(Z_n)| = \varphi(n)$, where $\varphi$ denotes the Euler totient. Thus $|N_{S_n}(\langle \tau \rangle)| = n\varphi(n)$.