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I'm self studying and came across the following problem that I'm having some trouble with.

If $G$ has order $p^2$, and $H$ is a subgroup of order $p$, then $H$ is normal in $G$.

I understand that this post shows a similar result. However, this is showing that $G$ contains some subgroup of order $p$ that is normal, not that the specific subgroup $H$ is normal. The concept of a center has not been formally introduced yet.

Here's my attempt.

Since $H$ has order $p$, it must be cyclic with generator $h$. Consider any $g \in G$. Let $a = g h g^{-1}$, and consider any $h^k \in H$. We want to show that $g h^k g^{-1} = (g h g^{-1})^k = a^k \in H$. Thus showing $\langle a \rangle = \langle h \rangle$ will prove the desired result.

Let $H_g = \langle h, a\rangle$. If $G \neq H_g$, then $H_g$ is a subgroup of order $p$ that contains both $a$ and $h$ so $H_g = \langle a \rangle = \langle h \rangle = H$. Thus $H$ is normal in $G$.

Now suppose $G = H_g$. In this case, we can write $g = a^{k_1} h^{k_2} \dots a^{k_m} $. I want maybe try to show by induction on the length of $g$ that $g \in H$.

Possibly on the right track? Any hints you can give? I've been struggling with this problem for a full week now.

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Nitin
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    This is not the elementary answer you are looking for, but for completeness here the high level proof: If $|G| = p^n$, then $|Z(G)|$ can neither be $1$ or $p^{n-1}$. For $n=2$ this implies $G=Z(G)$, i.e the group is abelian, which means all subgroups are normal. – Simon Oct 21 '16 at 18:05
  • That's quick slick, thanks for the answer. – Nitin Oct 21 '16 at 18:13
  • This is not remotely a resolution but I was looking at the latest edition of the text and it looks like of the problems in the section, this one (and only this one) was taken out. I'm giving up on this and moving on (forest for the trees and all that). – Nitin Oct 23 '16 at 06:23

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