Let $p$ be an odd prime. Prove that if $P$ is a non-cyclic $p$-group then $P$ contains a normal subgroup $U$ with $U\cong\Bbb Z_p\times\Bbb Z_p$.

Question

20. Let $p$ be an odd prime. Prove that if $P$ is a non-cyclic $p$-group then $P$ contains a normal subgroup $U$ with $U\cong\Bbb Z_p\times\Bbb Z_p$. (Abstract Algebra: Dummit & Foote, Semidirect Products)

The authors provide us with some hints. So I may as well incorporate them in the following attempt:

We proceed with induction on $|P|$. The cases are trivial for $|P|=p$ or $p^2$.
If $|P|=p^3$, $P\cong\Bbb Z_{p^2}\times\Bbb Z_p,$ or $\cong\Bbb Z_p\times\Bbb Z_p\times\Bbb Z_p$, or $\cong\Bbb Z_{p^2}\rtimes\Bbb Z_p$, or $\cong(\Bbb Z_p\times\Bbb Z_p)\rtimes\Bbb Z_p$.
For an abelian group, every subgroup is normal. Both $\Bbb Z_{p^2}\times\Bbb Z_p$ and $\Bbb Z_p\times\Bbb Z_p\times\Bbb Z_p$ obviously contain such subgroup $U$.

If $P\cong(\Bbb Z_p\times\Bbb Z_p)\rtimes\Bbb Z_p=\langle a,b,c$ s.t. $a^p=b^p=c^p=1,ab=ba,cac^{-1}=ab,cbc^{-1}=b\rangle$, take $U=\langle a,b\rangle$.

If $P\cong\Bbb Z_{p^2}\rtimes\Bbb Z_p=\langle d,e$ s.t. $d^{p^2}=e^p=1,ede^{-1}=d^{1+p}\rangle$, take $U=\langle d^p,e\rangle$. $d^p$ and $e$ commute because $ed^pe^{-1}=(ede^{-1})^p=(d^{1+p})^p=d^{p+p^2}=d^p$.

Now we really actually proceed with induction. But before that, notice that the result is true if $P$ is an abelian group: every subgroup is normal and it's non-cyclic so it contains two elementary divisors anyway. Write $P\cong\Bbb Z_{p^{\beta_1}}\times\Bbb Z_{p^{\beta_2}}\times\dots$. Find an element of order $p$ in the first factor and another in the second factor, they generate the $U$ we want.

Now the inductive hypothesis: suppose the result is true for any group with order $<|P|=p^\alpha$. Let $Z\le Z(P)$ with $|Z|=p$. Then we use this:

Lemma. if $P/Z$ is cyclic, then $P$ is abelian.

So Assume $P/Z$ is not cyclic. $|P/Z|<|P|$. Being a quotient of a $p$-group $P$, $P/Z$ is a $p$-group. By the inductive hypothesis, $P/Z$ contains a normal subgroup $H/Z\cong\Bbb Z_p\times\Bbb Z_p$. Let $\pi:P\rightarrow P/Z$ defined by $\pi:p\mapsto pZ$ be the natural projection homomorphism of the quotient. $\pi^{-1}(H/Z)=H$, the complete preimage of $H/Z$, is of order $p^3$.

Let $kerf=\{h\in H$ s.t. $h^p=1\}$ be the kernel of the following $p$-th power map. I also show some established results:

Let $p$ be an odd prime. $|H|=p^3$, the $p$-th power map of $H$, $f$, is defined as $f:h\mapsto h^p$. This is a homomorphism of $H$ into $Z(H)$. If $H$ is not cyclic, $|kerf|=p^2$ or $p^3$.

$H$ is not cyclic, because its quotient $H/Z$ is not cyclic. Also, $kerf\unlhd^{char}H$. This is because $kerf$ contains the identity and all the elements of $H$ that are of order $p$, and if $\sigma\in\operatorname{Aut}(H)$, $\sigma$ preserves orders of elements, so it permutes the elements of order $p$. So $\sigma(kerf)=kerf$.

We have $kerf\unlhd^{char}H\unlhd P$, i.e. $kerf\unlhd P$. And $kerf\cong\Bbb Z_p\times\Bbb Z_p$ (if so, then we're done), or $\cong\Bbb Z_p\times\Bbb Z_p\times\Bbb Z_p$, or $(\Bbb Z_p\times\Bbb Z_p)\rtimes\Bbb Z_p$.

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What if $kerf\cong\Bbb Z_p\times\Bbb Z_p\times\Bbb Z_p$, or $(\Bbb Z_p\times\Bbb Z_p)\rtimes\Bbb Z_p$? Surely for the former one, I cannot find characteristic subgroups of order $p^2$. There has to be some other way of constructing normal subgroup... For the latter one, is $\langle a,b\rangle$ characteristic? Why if it is?

Answer 1

Note that $H/Z$ has a subgroup $K/Z$ of order $p$ with $K$ normal in $G$. (That follows easily from the fact that $H/Z$ has nontrivial intersection with $Z(G/Z)$.)

If $K \cong C_p \times C_p$ then we are done. Otherwise $K \cong C_{p^2}$, and in that case $|\ker f| < p^3$, so we are not in either of the two cases that you have not been able to handle.

Answer 2

$P$ is a $p$-group so $\left|P\right|=p^{\alpha}$ for some $\alpha >1$ ($P$ is cyclic for $\alpha =1$). Theorem~4.3.8 implies a non-trivial center $Z\left(P\right)$ and $\left|Z\left(P\right)\right|=p^{\beta}$ for some $1\leq\beta\leq\alpha$. By Cauchy, the center contains a cyclic subgroup $Z_{p}$ of order $p$. Observe that as a subgroup of the center, this subgroup is normal in $P$. Then there is the factorization $P/Z_{p}$ and $\left|P/Z_{p}\right|=p^{\alpha -1}$. It is evident that a proof by induction is a possibility. Induct on the exponent $\alpha$. For $\alpha =2$, there is only the abelian possibility $Z_{p}\times Z_{p}$. Suppose $P/Z_{p}$ is cyclic so $P$ is both abelian (Exercise~3.1.36, it is an easy generalization to show that an identical proof holds for any subgroup of the center) and non-cyclic. Then all subgroups are normal and it is obvious that $P$ contains a normal subgroup $U\cong Z_{p}\times Z_{p}$ (Theorem~5.2.5).

On the other hand, suppose $P/Z_{p}$ is non-cyclic. The induction hypothesis provides a normal subgroup $Z_{p}\times Z_{p}$ with complete pre-image $H\unlhd P$ of order $\left|H\right|=p^{3}$. Let $H_{0}=\left\{\left. x\in H\right|\ x^{p}=1\right\}$ and consider any automorphism $\sigma\in\text{Aut}\left(H\right)$ so $1=\sigma\left(1\right)=\sigma\left(x^{p}\right)=\sigma\left(x\right)^{p}$ for all $x\in H_{0}$. By Exercise~5.4.9, $H_{0}$ is a characteristic subgroup of $H$ of order $p^{2}$ or $p^{3}$. Comparing to Example~5.5(7), $\left|H_{0}\right|=p^{2}$ obviously corresponds to the characteristic subgroup $\left\langle y^{p}\right\rangle\times\left\langle x\right\rangle$ of the semidirect product with presentation \begin{align*} \left\langle y\right\rangle\rtimes\left\langle x\right\rangle &=\left\langle\left. x,y\right|\ x^{p}=y^{p^{2}}=1,\ xyx^{-1}=y^{p+1}\right\rangle \end{align*} and since $\left\langle y^{p}\right\rangle\times\left\langle x\right\rangle\text{char}\left\langle y\right\rangle\rtimes\left\langle x\right\rangle\unlhd P$, it follows that $\left\langle y^{p}\right\rangle\times\left\langle x\right\rangle\unlhd P$.

The other possibility, $\left|H_{0}\right|=p^{3}$, obviously corresponds to the semidirect product with presentation \begin{align*} \left(\left\langle a\right\rangle\times\left\langle b\right\rangle\right)\rtimes\left\langle x\right\rangle &=\left\langle\left. x,a,b\right|\ x^{p}=a^{p}=b^{p}=1,\ ab=ba,\ xax^{-1}=ab,\ xbx^{-1}=b\right\rangle \end{align*} and it is not clear that the subgroup $\left\langle a\right\rangle\times\left\langle b\right\rangle$ is normal in $P$. However, it is obvious that $\left\langle b\right\rangle =Z\left(H_{0}\right)$ so $\left\langle b\right\rangle$ is normal in $P$ (the center is characteristic). Then $H_{0}/\left\langle b\right\rangle\cong\left\langle a\right\rangle\times\left\langle x\right\rangle$ so both $\left\langle a\right\rangle\times\left\langle b\right\rangle/\left\langle b\right\rangle$ and $\left\langle x\right\rangle\times\left\langle b\right\rangle/\left\langle b\right\rangle$ are normal in $H_{0}/\left\langle b\right\rangle$. In fact, the subgroups \begin{align*} \left\langle ax\right\rangle &\times\left\langle b\right\rangle/\left\langle b\right\rangle\\ \left\langle a^{2}x\right\rangle &\times\left\langle b\right\rangle/\left\langle b\right\rangle\\ &\ \,\vdots\\ \left\langle a^{p-1}x\right\rangle &\times\left\langle b\right\rangle/\left\langle b\right\rangle \end{align*} are normal in $H_{0}/\left\langle b\right\rangle$ as well so there are $p+1$ normal subgroups in $H_{0}/\left\langle b\right\rangle$ (see Example~5.1(3)). Then the exhaustive list of normal subgroups of order $p^{2}$ in $H_{0}$ is \begin{align*} \left\langle x\right\rangle &\times\left\langle b\right\rangle\\ \left\langle a\right\rangle &\times\left\langle b\right\rangle\\ \left\langle ax\right\rangle &\times\left\langle b\right\rangle\\ \left\langle a^{2}x\right\rangle &\times\left\langle b\right\rangle\\ &\ \,\vdots\\ \left\langle a^{p-1}x\right\rangle &\times\left\langle b\right\rangle \end{align*}

A simple example will suffice to establish the normal property for $\left\langle x\right\rangle\times\left\langle b\right\rangle$ in $P $. For $p=3$, let $H=\left\langle x\right\rangle$ and consider the representation obtained by left multiplication by the elements of $P$ on the $9$ left cosets \begin{align*} &\left\{H,aH,a^{2}H,bH,b^{2}H,abH,ab^{2}H,a^{2}bH,a^{2}b^{2}H\right\} \end{align*} Observe that $ax^{2}a^{-1}=ax^{-1}a^{-1}=x^{-1}xax^{-1}a^{-1}=x^{-1}aba^{-1}=x^{-1}b=bx^{2}$ so $H$ is not normal in $P$ and has no normal subgroups ($\left|H\right|=3$). Then the representation is faithful and $a$ and $b$ are represented as $3$ disjoint cycles of $3$ whereas $x$ is $2$ disjoint cycles of $3$. For the representation under discussion, \begin{align*} a &=\left(1\ 2\ 3\right)\left(4\ 6\ 8\right)\left(5\ 7\ 9\right)\\ b &=\left(1\ 4\ 5\right)\left(2\ 6\ 7\right)\left(3\ 8\ 9\right)\\ x &=\left(2\ 6\ 7\right)\left(3\ 9\ 8\right) \end{align*} and \begin{align*} ax &=\left(1\ 2\ 8\right)\left(3\ 5\ 7\right)\left(4\ 6\ 9\right)\\ a^{2}x &=\left(1\ 3\ 7\right)\left(2\ 4\ 8\right)\left(5\ 9\ 6\right) \end{align*} It follows that $\left\langle x\right\rangle\times\left\langle b\right\rangle$ is unique in $H_{0}$, hence characteristic in $H_{0}$ (Exercise~4.4.7) and normal in $P$. It is easy to generalize the latter argument for $p>3$. Thus the result.