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If $G$ is a group whose order is $p^n$($p$ is prime), then $G$ is solvable.

How am I going to show this? Any help is appreciated. Thank you.

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Hint: The center of such a group is non-trivial by the class formula. –  Tobias Kildetoft Mar 18 '13 at 17:52
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See Rotman's or Rose's books, if you have an accesses to them. –  B. S. Mar 18 '13 at 17:54
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1 Answer 1

up vote 5 down vote accepted

Try by induction on the power of $p$. If $n=1$, $G$ is solvable by definition as a cyclic group of prime order.

Suppose that statement is true for all $k\leq n-1$. Suppose $|G|=p^n$. By the class equation, the center $Z(G)$ is nontrivial. So $Z(G)$ is normal in $G$ and abelian, hence solvable.

So either $G/Z(G)$ is a $p$-group of smaller order, or it is trivial.

The key theorem to remember is that if $H\unlhd G$ and $H$ is solvable and $G/H$ is solvable, then $G$ is also solvable. If $|G/Z(G)|< p^n$, then by induction $G/Z(G)$ is solvable, so $G$ is solvable. Otherwise you just have $G=Z(G)$.

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Thanks Ben..... –  Philip Benj Marcoby Eragon Mar 18 '13 at 19:42
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