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I think my reasoning is wrong, but if the intersection only contains the identity, doesn't that imply that the only commutator in N is {e}, so doesn't that mean N is automatically commutative? Why was it necessary to state that N was a normal subgroup? Thanks!

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@DonAntonio : the derived subgroup of $S_3$ is $A_3$ –  Prahlad Vaidyanathan Apr 21 at 14:49
    
Oh, rats! Somehow I indeed read "derived group" but I understood "center of..." ! Thanks, deleting. –  DonAntonio Apr 21 at 14:55

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Your argument is correct to show that $N$ is abelian (whether or not $N$ is a normal subgroup). But to show that $N$ is in the centre of $G$ is stronger --- you have to show that every element of $N$ commutes with every element of $G$. This will require the assumption that $N$ is normal.

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For example $N=\{(), (1,2)\} \leq G=S_3 = \{ (), (1,2), (1,2,3), (2,3), (1,3,2), (1,3) \}$. $G'=\{(),(1,2,3),(1,3,2)\}$ and $N \cap G'=\{()\}$ so $N$ is abelian, but $Z(G)=\{()\}$ so $N$ is not a subgroup of the center. Indeed, $(1,2) \in N$ does not commute with any element of $G$ outside of $N$. –  Jack Schmidt Apr 21 at 15:15

(1) First, note that $\;N\lhd G\iff [N,G]\le N\;$

(2) Also, note that $\;x\in G\;$ is a central element iff $\;[x,G]=1\;$

So since $\;N\,,\,G'\lhd G\; $, we get

$$[N,G]\le[N,G']\le N\cap G'=1\iff N\le Z(G)$$

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I'm unfamiliar with the notations you used, do you mind letting me know what [N,G] means? Thanks! –  Rod Apr 21 at 15:06
    
@Rod : $$[N,G]:=\langle,[n,g]\;;\;n\in N\,,\,g\in G\;\rangle =$$the subgroup generated by all the commutators of the form $\;[n,g]\;$ ... –  DonAntonio Apr 21 at 15:09

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