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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)?

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if a is correct so d is correct as well. –  B. S. Jan 26 '13 at 11:31
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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
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2 Answers 2

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.

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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 –  B. S. Feb 6 '13 at 2:50
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Hint: Try to show that every non-abelian $p-$ group $G$ of order $p^3$ has this property: $$Z(G)\cong G'\cong\mathbb Z/p\Bbb{Z}$$

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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. –  B. S. Jan 26 '13 at 15:21
    
+1 Well-chosen hint, and follow-up! –  amWhy Jan 27 '13 at 15:15
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