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Let $ G$ be a group of order $16$. Show that $G$ must contain a normal subgroup $H$ of order $4$.

I tried the Sylow first theorem, that is $\{e\}\triangleleft H_1\triangleleft H_2\triangleleft H_3\triangleleft G $ where $|H_i|=2^i$. Then I wanted to prove that the normalizer of $H_2$ is $G$. But I couldn't, maybe it's not the correct way to prove it!

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Some MathJax advice: use |, or \lvert and \rvert, for "absolute value" bars. The lines produced by \mid have extra space on either side. –  Zev Chonoles Aug 15 '13 at 0:13
    
I would recommend trying to prove the more general statement that any group of order $p^n$ (for $p$ prime) has a normal subgroup of order $p^m$ for all $0 \le m \le n$. You can do it by induction on $n$, using the nontriviality of the centre of a $p$-group. –  Derek Holt Aug 15 '13 at 9:23

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If $|Z(G)| \geq 4$ we are done. Otherwise $|Z(G)| = 2$ since $G$ is a $p$-group. Any group of order $8$ has a normal subgroup of order $2$ (so, of index 4), therefore considering the quotient $G/Z(G)$, $G$ must have a normal subgroup of index $4$.

Actually more is true: Normal subgroups of p-groups

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Why when $|Z(G)|=8$ we are easily done? –  Ronald Aug 15 '13 at 9:17
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@Danial Because $Z(G)$ is abelian of order $8$. (So $Z(G)$ is isomorphic to one of $C_8$, $C_4\times C_2$ or $C_2\times C_2\times C_2$. All contain a subgroup of order $4$.) –  user1729 Aug 15 '13 at 10:34

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