Prove that every group $G$ with $p^n$ ($n\ge4$) elements and center with $p$ elements has an abelian subgroup of order $p^3$

Question

I'm new in this forum so I hope I haven't made any mistake. I have to prove the above assertion. I've already proven that every such $G$ has an element $x$ whose centralizer has $p^{n-1}$ elements. I tried to prove it by induction on $n$. I proved the case $n=4$ where the group of order $p^3$ is the centralizer of one of the elements. Then supposing it is true for $n=k$ I want to prove it for $n=k+1$. I know once again that $G$ contains an element whose centralizer has $p^{k+1-1}=p^k$ elements. However this subgroup obviously doesn't have $p$ elements in the center since there are at least $p+1$ elements in it (the element itself and the center of $G$ which has $p$ elements) and so I can't apply the induction. I'm sorry for my English also, it's not my first language. If anybody has any suggestion I would be really grateful. Thank you in advance.

Answer 1

You already have it! If you know it holds when $G$ has order $p^4$ and center of order $p$ (the hardest part of the problem), then it follows easily for every other group simply by taking any subgroup of order $p^4$ of your group $G$.

I give a full proof for completeness:

Theorem. If $G$ has order $p^4$, then $G$ contains an abelian subgroup of order $p^3$.

Proof. If $G$ is abelian, this is trivial.

If $|Z(G)|=p^2$, then let $H$ be a subgroup of $G$ of order $p^3$ containing $Z(G)$. Then $H$ must be abelian, since $Z(G)\leq Z(H)$ and $[H:Z(G)]=p$.

Now assume that $|Z(G)|=p$, $x\in Z(G)$ with $x\neq e$. Let $y\in G$ be such that $yZ(G)\in Z(G/Z(G))$ has order $p$. Then $\langle yZ(G)\rangle\triangleleft G/Z(G)$, and therefore $\langle x,y\rangle\triangleleft G$ has order $p^2$ and is abelian. We have two cases:

1. If $\langle x,y\rangle$ is cyclic of order $p^2$, generated by $z$, then its automorphism group has order $p(p-1)$. The action of $G$ on $\langle z\rangle$ by conjugation induces a morphism $G\to \mathrm{Aut}(\langle z\rangle)$ which has nontrivial kernel of order at least $p^3$ (since the image has order dividing $p$), and this kernel is $C_G(z)$. Since $z$ is not central (one of its powers is $y$ which is not central), $C_G(z)$ has order exactly $p^3$. It has a central cyclic subgroup of order $p^2$, and thus is abelian or order $p^3$.

2. If $\langle x,y\rangle\cong C_p\times C_p$, then the automorphism group is isomorphic to $\mathsf{GL}(2,p)$, which has order $(p^2-1)(p^2-p)= p(p-1)^2(p+1)$. So again the action of $G$ on $\langle x,y \rangle = C_p\times C_p$ induces a map into $\mathsf{GL}(2,p)$ which has image of order dividing $p$; thus, the centralizer of $\langle x,y\rangle$ has order at least $p^3$. Since $\langle x,y\rangle$ is not central, the centralizer of the subgroup is of order exactly $p^3$, and contains a central subgroup of order $p^2$, thus again we get that it is abelian of order $p^3$.

This proves the theorem. $\Box$

Corollary. If $G$ has order $p^n$, $n\geq 4$, then $G$ has an abelian subgroup of order $p^3$.

Proof. Let $G$ be a group of order $p^n$ with $n\geq 4$ (by Sylow's First Theorem, any group of order $p^i$ whose index is a multiple of $p$ is contained in a subgroup of order $p^{i+1}$; in particular, a group of order $p^n$ has subgroups of order $p^j$ for any $0\leq j\leq n$). Let $H$ be a subgroup of $G$ of order $p^4$. Then $H$ contains an abelian subgroup of order $p^3$, and therefore so does $G$. $\Box$