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

Let $G$ be a group with 81 elements and $H$ a subgroup with 27 elements. Which of the following is not true?

  • (a) $H$ is a normal subgroup of $G$
  • (b) $Z(H)\neq 1$
  • (c) $H'=1$
  • (d) $G'\subseteq H$

I know that (b) is true. Since the center of every nontrivial p-group is nontrivial (p is prime). Also, (d) is true since $G/H$ has 3 elements and hence $G/H$ is abelian. So $G'\subseteq H$. But (a) and (c)?

share|cite|improve this question
if a is correct so d is correct as well. – Babak S. Jan 26 '13 at 11:31
How can you say for (d) that $G/H$ is an abelian group if you don't know whether (a) is true? – Hagen von Eitzen Jan 26 '13 at 11:33
@HagenvonEitzen For subgroup $H$ of $G$, $G'\subseteq H$ if and only if $G/H$ be abelian – aliakbar Jan 26 '13 at 11:36
The point of @HagenvonEitzen was that you need to know $H$ is normal to be able to talk about the quotient group $G/H$. Or, to rephrase your statement, for $H$ a subgroup of $G$, one has $G' \subseteq H$ if and only if $H$ is a normal subgroup of $G$ and $G/H$ is abelian, – Andreas Caranti Jan 26 '13 at 11:41
up vote 6 down vote accepted

(a) is also true, for instance because a subgroup of a finite group whose index is the smallest prime dividing the order of the group is normal.

You may also want to see why (c) fails. You need to know an example of a nonabelian group of order 27, and then take the direct product with a group of order 3.

share|cite|improve this answer
Thank you very much – aliakbar Jan 26 '13 at 11:49
You're welcome. – Andreas Caranti Jan 26 '13 at 11:53
Sorry Prof., May I ask you to have a look at my approach. Thanks – Babak S. Feb 6 '13 at 2:50

Hint: Try to show that every non-abelian $p-$ group $G$ of order $p^3$ has this property: $$Z(G)\cong G'\cong\mathbb Z/p\Bbb{Z}$$

share|cite|improve this answer
Please explain for me by details. – aliakbar Jan 26 '13 at 11:38
@aliakbar: When $|G|=p^3$ and $G$ is non-abelian so $|G/Z(G)|=p, p^2, 1$. If $|G/Z(G)|=p$ then $G/Z(G)$ would be cyclic and therefore $G$ be abelian. If $|G/Z(G)|=1$ then $G\cong Z(G)$ and again $G$ be abelian.So $|G/Z(G)|=p^2$ and therefore $|Z(G)|=p$ and $G'\subset Z(G)$. So the option c would be your option. – Babak S. Jan 26 '13 at 15:21
+1 Well-chosen hint, and follow-up! – amWhy Jan 27 '13 at 15:15

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.