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I'm trying to show that a normal subgroup $N$ of a p-group $G$ intersects $Z(G)$ nontrivially (please don't tell how to show it), but it seem it is quite a trivial question considering the following argument:

Consider $Z(N)$, a nontrivial subgroup of $N$. Since any element in $Z(G)$ also commutes with the elements of $N$, $Z(G) \subseteq Z(N)$, but we know that $Z(N) \leq N$, so $ \{e\} \not = Z(G) \subseteq N.$

If the answer was this, I don't think my Algebra professor would ask it, so what is wrong with the above argument ?

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  • $\begingroup$ It's certainly not the case that $Z(G)\subseteq Z(N)$. $\endgroup$ – Lord Shark the Unknown Dec 15 '18 at 6:08
  • $\begingroup$ @LordSharktheUnknown Why ? $\endgroup$ – onurcanbektas Dec 15 '18 at 6:08
  • $\begingroup$ @LordSharktheUnknown By definition, any element that commutes with all the elements in the group shouldn't commute with the elements in the subgroup ? $\endgroup$ – onurcanbektas Dec 15 '18 at 6:09
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    $\begingroup$ What if $N$ is a proper subgroup of $Z(G)$? $\endgroup$ – Lord Shark the Unknown Dec 15 '18 at 6:13
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    $\begingroup$ An element $g$ of $G$ belongs to $Z(N)$ if and only if it commutes with all the elements of $N$ and it belongs to $N$. If the latter condition is not verified you only know that $g \in C_G(N)$ which is in general bigger than $Z(N)$. $\endgroup$ – Pietro Gheri Dec 15 '18 at 12:40
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$Z(N)$ and $Z(G)$ need not be at all related. If $N$ is abelian, then $Z(N) =N$, but the center of $G$ might intersect $N$ trivially. For example, the center of $S_3\times \mathbb Z_3$ intersects a subgroup of order $2$ in the first factor trivially.

In general, $Z(G) $ is a subgroup of the centralizer of $N$, but not necessarily the center.

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  • $\begingroup$ In my notation, $Z(G):= C(G)$. $\endgroup$ – onurcanbektas Dec 15 '18 at 14:49
  • $\begingroup$ @onur Yes, but $Z(N) \neq C(N) $ in general, unless you're using highly unusual notation. $\endgroup$ – Matt Samuel Dec 15 '18 at 14:51
  • $\begingroup$ what is $Z(N)$ in your notation ? $\endgroup$ – onurcanbektas Dec 15 '18 at 14:53
  • $\begingroup$ @onur The set of all elements of $N$ that commute with every element of $N$. $C(N) $ is the set of all elements of $G$ that commute with every element of $N$. $\endgroup$ – Matt Samuel Dec 15 '18 at 14:56
  • $\begingroup$ Oh, I see. Thanks both for the answer and the replies. $\endgroup$ – onurcanbektas Dec 15 '18 at 14:57

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