1
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • $\begingroup$ math.overflow is for research-level questions. You may want to look at math.stackexchange instead, but please read the FAQ there before posting. $\endgroup$
    – Arturo Magidin
    Dec 12 at 21:52
  • 2
    $\begingroup$ I am not an MSE regular, but, given that this problem describes your attempts to solve the problem, it seems to me that it could be well received there. $\endgroup$
    – LSpice
    Dec 12 at 22:15
  • $\begingroup$ Hint: (0) check that it's enough to deal with G of order p^4 (1) prove that there is a normal subgroup H of order p^2 (2) show that for every group W of order p^2, the p-Sylow of Aut(W) have order p. (3) Deduce that the centralizer of H has order p^3 or p^4 and conclude. $\endgroup$
    – YCor
    Dec 12 at 23:02

0

Reset to default

Browse other questions tagged or ask your own question.