If GG has order p2p^2, and HH is a subgroup of order pp, then HH is normal (Searching for cleaner proof)

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

If GG has order p2p^2, and HH is a subgroup of order pp, then HH is normal in GG.

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

Here’s my attempt.

Since HH has order pp, it must be cyclic with generator hh. Consider any g∈Gg \in G. Let a=ghg−1a = g h g^{-1}, and consider any hk∈Hh^k \in H. We want to show that ghkg−1=(ghg−1)k=ak∈Hg h^k g^{-1} = (g h g^{-1})^k = a^k \in H. Thus showing ⟨a⟩=⟨h⟩\langle a \rangle = \langle h \rangle will prove the desired result.

Let Hg=⟨h,a⟩H_g = \langle h, a\rangle. If G≠HgG \neq H_g, then HgH_g is a subgroup of order pp that contains both aa and hh so Hg=⟨a⟩=⟨h⟩=HH_g = \langle a \rangle = \langle h \rangle = H. Thus HH is normal in GG.

Now suppose G=HgG = H_g. In this case, we can write g=ak1hk2…akmg = a^{k_1} h^{k_2} \dots a^{k_m} . I want maybe try to show by induction on the length of gg that g∈Hg \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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This is not the elementary answer you are looking for, but for completeness here the high level proof: If |G|=pn|G| = p^n, then |Z(G)||Z(G)| can neither be 11 or pn−1p^{n-1}. For n=2n=2 this implies G=Z(G)G=Z(G), i.e the group is abelian, which means all subgroups are normal.
– Simon
2 days ago

  

 

That’s quick slick, thanks for the answer.
– Nitin
2 days ago

  

 

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
21 hours ago

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