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Do the definition of the center of a group apply to subgroups. For example if $N$ is a normal subgroup of $G$, I want to consider $Z(N) = \{ n \in N |\space nx = xn \space\space x \in N\} $. I realize that $Z(N)$ would be a subset of $C(N)$.

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The subgroup is a group, after all -- why shouldn't you be able to take its center? –  mike4ty4 Oct 6 '12 at 0:43
    
I haven't seen it used in that way. I just wanted to make sure before I apply a theorem using centers of groups. –  Simply Brian Oct 6 '12 at 0:47
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Sure, you laid out the definition right there. $N$ doesn't have to be normal, either. You can take the center of any subgroup.

Something interesting to note is that, if $N\leqslant G$, it is not impossible that $Z(N)$ is bigger than $Z(G)$. For example, if you've got the group $D_n$, where $n$ is even and $r$ is the generating "rotation" of order $n$, the center of the subgroup $\langle r \rangle \leqslant D_n$ is $Z(\langle r \rangle) = \langle r \rangle$, yet $Z(D_n)=\langle r^{n/2} \rangle$. Of course, this isn't always the case: for example, if $G=D_n \times C_2$, we have $D_n\times 1\unlhd G$. In this case, $Z(D_n\times 1)=\langle r^{n/2}\rangle \times 1\leqslant Z(G)=\langle r^{n/2}\rangle \times C_2$.

You can also have a subgroup $N\leqslant G$ where $Z(N)>1$, $Z(G)>1$, and yet $Z(N)\cap Z(G)=1$ - in fact, there is an example of this in $D_n$ if you care to find it.

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