Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Hint: The center of such a group is non-trivial by the class formula. – Tobias Kildetoft Mar 18 '13 at 17:52
See Rotman's or Rose's books, if you have an accesses to them. – Babak S. Mar 18 '13 at 17:54
up vote 9 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 $G=Z(G)$.

share|cite|improve this answer
Thanks Ben..... – Philip Benj Marcoby Eragon Mar 18 '13 at 19:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.