# Let $p$ be an odd prime. A non-cyclic $p$-group contains a normal subgroup isomorphic to the elementary abelian group of order $p^2$.

1. 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 $(\Bbb Z_p\times\Bbb Z_p)\rtimes\Bbb Z_p$.

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?

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.

• How to see that $(H/Z)\cap Z(P/Z)\neq 1$? – user441558 Aug 1 '17 at 0:05
• Never mind... I just found this result in the next chapter: a nontrivial normal subgroup of a $p$-group intersects the center nontrivially... – user441558 Aug 1 '17 at 7:15