I'm new in this forum so I hope I haven't made any mistake. I have to prove what I've written in the title. 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.

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

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.

I'm new in this forum so I hope I haven't made any mistake. I have to prove what I've written in the title. 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.

I'm new in this forum so I hope I haven't made any mistake. I have to prove what I've written in the title. 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.