Suppose we have a prime number $p$ as well as a positive integer $n$. Consider a group $G$ with $p^n$ elements. I want to prove that $G$ has a normal subgroup with $p^{n-1}$ elements. I found this in the book I am reading. When I think about it, it makes sense, but I was a little iffy on how to prove it.
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1Hint: Such a group has a non-trivial center. Take the quotient by the center and apply induction. – Tobias Kildetoft Dec 05 '13 at 19:59
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Are you sure that is a duplicate. It looks like they showed it for $p^{n+1}$ on that one – user113561 Dec 05 '13 at 20:19
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@user113561 The proof in the other question is more general. It shows that there is a normal subgroup of order $p^k$ for all $0 \le k \le n$, not just $p^{n-1}$. – Ayman Hourieh Dec 06 '13 at 00:51
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HINT. First, use the class formula $$ |G|=|Z(G)|+\sum_{a \in G,\; a \notin Z(G)} |G:C(a)| $$ to show that $|Z(G)|>1$ (that is the center is nontrivial), where the $a \in G$ are distinct representatives for the conjugacy classes of $a \in G$.
Finally, use induction on $k$ to show that $G$ has a normal subgroup of order $p^k$ for $0 \le k \le n$.
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