## Tag Archives: normal subgroup

### A finite group whose order is the product of three distinct primes has a normal Sylow subgroup of largest order

Let $G$ be a group of order $pqr$ where $p < q < r$ and $p$, $q$, and $r$ are primes. Prove that a Sylow $r$-subgroup of $G$ is normal.

Recall that some Sylow subgroup of $G$ is normal. Let $P$, $Q$, and $R$ denote Sylow $p$-, $q$-, and $r$-subgroups of $G$, respectively.

Suppose $n_p(G) = 1$. Note that $n_r(G/P) \in \{1,q\}$, and since $q < r$, $n_r(G/P) = 1$. Let (via the lattice isomorphism theorem) $\overline{R} \leq G$ be the subgroup whose quotient is the unique Sylow $r$-subgroup in $G/P$; we have $|\overline{R}| = pr$, and $P \leq \overline{R}$ is normal. Moreover, because $p < r$, $\overline{R}$ has a unique Sylow $r$-subgroup $R$, and we have $\overline{R} \cong P \times R$. Now if $R^\prime \leq G$ is a Sylow $r$-subgroup, then $R^\prime P/P = \overline{R}/P$, so that $\overline{R} = R^\prime P$. Since $R^\prime \leq \overline{R}$ is Sylow, $R^\prime = R$. Thus $n_r(G) = 1$.

The same argument works if $n_q(G) = 1$.

### Every group of order 36 has a normal Sylow subgroup

Prove that if $G$ is a group of order 36, then $G$ has either a normal Sylow 2-subgroup or a normal Sylow 3-subgroup.

Note that $36 = 2^2 \cdot 3^2$, so that Sylow’s Theorem forces $n_2(G) \in \{1,3,9\}$ and $n_3(G) \in \{1,4\}$.

Suppose $n_3(G) \neq 1$. Then $n_3(G) = 4 \not\equiv 1$ mod 9, so that by Lemma 13 and this previous exercise, there exist $P_3,Q_3 \in \mathsf{Syl}_3(G)$ such that $P_3 \cap Q_3$ is nontrivial. Consider now $C_G(P_3 \cap Q_3)$; since $P_3$ and $Q_3$ are abelian, $\langle P_3,Q_3 \rangle$ centralizes $P_3 \cap Q_3$, and moreover we can see by Sylow’s Theorem that $G = \langle P_3, Q_3 \rangle$. Thus $P_3 \cap Q_3 \leq Z(G)$, and in fact $P_3 \cap Q_3$ is the intersection of all Sylow 3-subgroups of $G$. Thus $G$ contains $2 + 6 + 6 + 6 + 6 = 26$ elements of 3-power order. Let $P_2 \leq G$ be a Sylow 2-subgroup; $P_2$ contains three nonidentity elements, whose products with the nonidentity elements in $P_3 \cap Q_3$ yield 6 elements of order 6 or 12. There are 4 elements left, which necessarily comprise a unique Sylow 2-subgroup.

### No simple groups of order 144, 525, 2025, or 3159 exist

Prove that there are no simple groups of order 144, 525, 2025, or 3159.

1. Note that $144 = 2^4 \cdot 3^2$. Let $G$ be a simple group of order 144. By Sylow’s Theorem, we have $n_2 \in \{1,3,9\}$ and $n_3 \in \{1,4,16\}$. Note that $|G|$ does not divide $5!$ since the largest power of 2 which divides $5!$ is $2^3$. Thus no proper subgroup of $G$ has index less than or equal to 5; in particular, $n_2(G) = 9$ and $n_3(G) = 16$. Note that $n_3 \not\equiv 1$ mod 9; by Lemma 13, there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $P_3 \cap Q_3$ is normal in $P_3$ and $Q_3$. In particular, $3^2$ divides $|N_G(P_3 \cap Q_3)|$, and since $N_G(P_3 \cap Q_3)$ contains more than one Sylow 3-subgroup (namely $P_3$ and $Q_3$) its order is divisible by 2. In particular, $[G : N_G(P_3 \cap Q_3)] \in \{1,2,4,8\}$. Since $G$ is simple and has no proper subgroups of index less than 6, in fact $[G : N_G(P_3 \cap Q_3)] = 8$, and we have $|N_G(P_3 \cap Q_3)| = 2 \cdot 3^2$.

However, note that this implies that $n_3(N_G(P_3 \cap Q_3)) = 1$, by Sylow’s Theorem, but that we know $N_G(P_3 \cap Q_3)$ contains at least two Sylow 3-subgroups. Thus we have a contradiction.

2. Note that $525 = 3 \cdot 5^2 \cdot 7$. Let $G$ be a simple group of order 525. By Sylow’s Theorem, we have $n_3 \in \{1,7,25,175\}$, $n_5 \in \{1,21\}$, and $n_7 \in \{1,15\}$. Note that $|G|$ does not divide $9!$ since the highest power of 5 dividing $9!$ is $5^1$. Thus $G$ has no proper subgroups of index at most 9. Moreover, $n_5(G) = 21 \not\equiv 1$ mod 25, so that by a previous exercise and Lemma 13, there exist $P_5, Q_5 \in \mathsf{Syl}_5(G)$ such that $P_5 \cap Q_5 \neq 1$ and $N_G(P_5 \cap Q_5)$ is divisible by $5^2$ and some other prime.

If $|N_G(P_5 \cap Q_5)| \in \{ 3 \cdot 5^2, 7 \cdot 5^2\}$, then $G$ has a proper subgroup of index less than 9, a contradiction. If $|N_G(P_5 \cap Q_5)| = 3 \cdot 5^2 \cdot 7$, then $P_5 \cap Q_5$ is normal in $G$, a contradiction.

3. Note that $2025 = 3^4 \cdot 5^2$. Let $G$ be a simple group of order 2025. By Sylow’s Theorem, we have $n_3(G) = 25$ and $n_5(G) = 81$. Note that $|G|$ does not divide $9!$, since the highest power of 5 dividing $9!$ is $5^1$.

Now since $n_3(G) = 25 \not\equiv 1$ mod 9, by Lemma 13 and a previous exercise there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^4$ and 5. But then $N_G(P_3 \cap Q_3)$ is either all of $G$ or has index 5; in either case we have a contradiction.

4. Note that $3159 = 3^5 \cdot 13$. Let $G$ be a simple group of order 3159. By Sylow’s Theorem we have $n_3(G) = 13$ and $n_{13}(G) = 27$. Note that $n_3(G) = 13 \not\equiv 1$ mod 9; thus by Lemma 13 and a previous exercise, there exist $P_3, Q_3 \in \mathsf{Syl}_3(G)$ such that $|N_G(P_3 \cap Q_3)|$ is divisible by $3^5$ and 13- thus $N_G(P_3 \cap Q_3) = G$, a contradiction.

### Every minimal normal subgroup of a finite solvable group is elementary abelian

For any group $G$ a minimal normal subgroup is a normal subgroup $M$ of $G$ such that the only normal subgroups of $G$ which are contained in $M$ are $M$ and 1. Prove that every minimal normal subgroup of a finite solvable group is an elementary abelian $p$-group for some prime $p$.

Let $G$ be a finite solvable group and let $N \leq G$ be a minimal normal subgroup.

If $N=1$, then the conclusion is trivial. Suppose $N \neq 1$.

Since $G$ is solvable, $N$ is solvable. Since $N \neq 1$, $N^\prime < N$ is proper; now $N^\prime$ is characteristic in $N$, hence normal in $G$. Since $N$ is minimal, we have $N^\prime = 1$; thus $N$ is abelian.

Recall from this previous exercise that $N_p = \{ x \in N \ |\ x^p = 1 \}$ is characteristic in $N$, hence normal in $G$, for all primes $p$. Thus $N_p \in \{1,N\}$ for all primes $p$. By Cauchy, some element of $N$ has prime order, so that $N_p \neq 1$ for some prime $p$; in particular $N_p = N$. Thus $N$ is an elementary abelian $p$-group.

### The Frattini subgroup of a finite group is nilpotent

Let $G$ be a finite group. Prove that $\Phi(G)$ is nilpotent.

We will use Frattini’s Argument to show that every Sylow subgroup is normal.

Let $P \leq \Phi(G)$ be a Sylow subgroup. By Frattini’s Argument, $G = \Phi(G) N_G(P)$. By a lemma to a previous theorem, $N_G(P)$ cannot be proper, so that $N_G(P) = G$ and we have $P \leq G$ normal. Then $P \leq \Phi(G)$ is normal; thus all of the Sylow subgroups of $\Phi(G)$ are normal, and thus $\Phi(G)$ is nilpotent.

### The Frattini subgroup is inclusion-monotone on normal subgroups

When $G$ is a finite group prove that if $N \leq G$ is normal, then $\Phi(N) \leq \Phi(G)$. Give an explicit example when this containment does not hold if $N$ is not normal in $G$.

[With help from the GroupProps wiki.]

We begin with some lemmas.

Lemma 1: Let $G$ be a group, and let $A,B,C \leq G$ with $A \leq C$ be subgroups. Then $A(B \cap C) = AB \cap C$. Proof: $(\subseteq)$ Let $x \in A(B \cap C)$. Then $x = ad$ where $d \in B$ and $d \in C$, so that $x \in AB \cap AC = AB \cap C$. $(\supseteq)$ Let $x \in AB \cap C$. Then we have $x = ab = c$ for some $a \in A$, $b \in B$, and $c \in C$. Then $b = a^{-1}c \in C$, so in fact $b \in B \cap C$. Thus $x \in A(B \cap C)$. $\square$

Lemma 2: Let $G$ be a finite group. If $H < G$ is a proper subgroup, then $\Phi(G)H < G$ is proper. Proof: $\Phi(G) \leq G$ is characteristic, hence normal, in $G$, so that $\Phi(G)H \leq G$ is a subgroup. Now $H$ is contained in some maximal subgroup $M \leq G$ since $G$ is finite, and $\Phi(G) \leq M$ by definition. Thus $\Phi(G)H \leq M < G$ is proper. $\square$

Now to the main result.

Let $N \leq G$ be normal and suppose by way of contradiction that $\Phi(N) \not\leq M$ for some maximal subgroup $M \leq G$. Since $\Phi(N) \leq N$ is characteristic and $N \leq G$ is normal, $\Phi(N) \leq G$ is normal. Thus $\Phi(N)M \leq G$ is a subgroup, and in fact $\Phi(N)M = G$.

Now $N = G \cap N = \Phi(N)M \cap N$ $= \phi(N)(M \cap N)$. However, since $\Phi(N) \not\leq M$, then $N \not\leq M$, so that $M \cap N \leq N$ is proper. But this implies (by Lemma 2) that $N < N$ is proper, a contradiction. Thus no such subgroup $M$ exists, and in fact $\Phi(N) \leq M$ for all maximal subgroups $M \leq G$. Hence $\Phi(N) \leq \Phi(G)$.

Now before we produce a counterexample, we prove another lemma.

Lemma 3: $\Phi(S_5) = 1$. Proof: $\Phi(S_5)$ is a proper normal subgroup, so that $\Phi(S_5) \in \{1, A_5\}$. Now $S_5$ contains a subgroup $H$ isomorphic to $S_4$; by Lagrange, we have that $[S_5,H] = 5$ is prime, so that $H$ is maximal. Recall also that $A_5$ is maximal; thus $|\Phi(S_5)|$ divides both $|H| = 24$ and $|A_5| = 60$; hence $|\Phi(S_5)|$ divides 12, and we have $\Phi(S_5) = 1$. $\square$

Consider $K = \langle (1\ 2\ 3\ 4) \rangle \leq S_5$; this subgroup is cyclic of order 4, and thus has a unique maximal subgroup which is isomorphic to $Z_2$. Thus $K \leq S_5$ but $\Phi(K) \not\leq \Phi(S_5)$.

### Some basic properties of nilpotent groups

Prove the following for $G$ an infinite nilpotent group.

1. Let $G$ be a nilpotent group. If $H \leq G$ is a nontrivial normal subgroup, then $H \cap Z(G)$ is nontrivial. In particular, every normal subgroup of prime order is in the center.
2. Let $G$ be a nilpotent group. If $H < G$ is a proper subgroup, then $H < N_G(H)$ is a proper subgroup.

First we prove some lemmas.

Lemma 1: Let $G$ be a group, and let $Z_k(G)$ denote the $k$-th term in the upper central series of $G$. Then $Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G)$. Proof: We proceed by induction on $k$. For the base case, if $k = 0$, we have $Z_0(G/Z(G)) = 1$ $= Z(G)/Z(G)$ $= Z_1(G)/Z(G)$. For the inductive step, suppose $Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G)$ for some $k \geq 0$. We have the following chain of equivalent statements.

 $x Z(G) \in Z_{k+1}(G/Z(G))$ $\Leftrightarrow$ $(xZ(G))Z_k(G/Z(G)) \in Z((G/Z(G))/Z_k(G/Z(G)))$ $\Leftrightarrow$ For all $yZ(G) \in G/Z(G)$, $(xyZ(G))Z_k(G/Z(G)) = (yxZ(G))Z_k(G/Z(G))$ $\Leftrightarrow$ For all $yZ(G) \in G/Z(G)$, $[x,y]Z(G) \in Z_k(G/Z(G))$ $\Leftrightarrow$ For all $y \in G$, $[x,y]Z(G) \in Z_{k+1}(G)/Z(G)$ $\Leftrightarrow$ For all $y \in G$, $[x,y] \in Z_{k+1}(G)$ $\Leftrightarrow$ For all $y \in G$, $xy Z_{k+1}(G) = yx Z_{k+1}(G)$ $\Leftrightarrow$ $x Z_{k+1}(G) \in Z(G/Z_{k+1}(G))$ $\Leftrightarrow$ $x Z_{k+1}(G) \in Z_{k+2}(G)/Z_{k+1}(G)$ $\Leftrightarrow$ $x \in Z_{k+2}(G)$ $\Leftrightarrow$ $xZ(G) \in Z_{k+2}(G)/Z(G)$

Thus $Z_{k+1}(G/Z(G)) = Z_{k+2}(G)/Z(G)$, and the conclusion holds for all $k$. $\square$

Lemma 2: Let $G$ be a group. If $G$ is nilpotent of class $k+1$, then $G/Z(G)$ is nilpotent of class $k$. Proof: $G/Z(G)$ is nilpotent. Using Lemma 1, we have $Z_k(G/Z(G)) = Z_{k+1}(G)/Z(G)$ $= G/Z(G)$, so the nilpotence class of $G/Z(G)$ is at most $k$. Now for $t < k$, we have $Z_t(G/Z(G)) = Z_{t+1}(G)/Z(G)$; since $G$ has nilpotence class $k+1$, $Z_{t+1}(G) \neq G$, and $Z_t(G/Z(G)) \neq G/Z(G)$. Thus the nilpotence class of $G/Z(G)$ is precisely $k$. $\square$

We now move to the main result.

1. We proceed by induction on the nilpotence class of $G$.

For the base case, if $G$ has nilpotence class $k =1$, then $G$ is abelian. Now any nontrivial normal subgroup $H \leq G$ satisfies $H \cap Z(G) = H \neq 1$.

For the inductive step, suppose the conclusion holds for any nilpotent group of nilpotence class $k$. $G$ be a nilpotent group of class $k+1$, and let $H \leq G$ be a nontrivial normal subgroup. Suppose by way of contradiction that $H \cap Z(G) = 1$; now consider the internal direct product $HZ(G) \leq G$. We have $HZ(G)/Z(G) \leq G/Z(G)$; moreover, since $H$ and $Z(G)$ are normal, $HZ(G)/Z(G) \leq G/Z(G)$ is normal. By Lemma 2, $G/Z(G)$ is nilpotent of nilpotence class $k$, so that, by the induction hypothesis, $HZ(G)/Z(G) \cap Z(G/Z(G))$ is nontrivial. Let $xzZ(G) \in HZ(G)/Z(G) \cap Z(G/Z(G))$, where $x \in H$ and $z \in Z(G)$. In fact, we have $xZ(G) \in Z(G/Z(G))$. Let $g \in G$ be arbitrary. Since $H$ is normal in $G$, $g^{-1}hg \in H$. Now $g^{-1}hgZ(G) = hZ(G)$ since $hZ(G)$ is central in $G/Z(G)$, so that $h^{-1}g^{-1}hg \in Z(G)$. Recall, however, that $H \cap Z(G) = 1$, so that $h^{-1}g^{-1}hg = 1$, and we have $gh = hg$. Thus $h \in Z(G)$, a contradiction. So $H \cap Z(G) \neq 1$.

2. We proceed again by induction on the nilpotence class of $G$.

For the base case, if $G$ has nilpotence class $k = 1$, then $G$ is abelian. Thus $N_G(H) = G$ for all subgroups $H$, and if $H < G$ is proper, then $H < N_G(H)$ is proper.

For the inductive step, suppose every nilpotent group of nilpotence class $k \geq 1$ has the desired property. Let $G$ be a nilpotent group of nilpotence class $k+1$, and let $H < G$ be a proper subgroup. Now $G/Z(G)$ is nilpotent of class $k$. Suppose $Z(G) \not\leq H$; then, since $Z(G) \leq N_G(H)$, $H \leq \langle H, Z(G) \rangle$ is proper, and $H \leq N_G(H)$ is proper. If $Z(G) \leq H$, then $H/Z(G) \leq G/Z(G)$ is proper. By the induction hypothesis, $H/Z(G) \leq N_{G/Z(G)}(H/Z(G)) = N_G(H)/Z(G)$ is proper, so that $H \leq N_G(H)$ is proper.

### 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.

### 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.