Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

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

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.

share|cite|improve this answer
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 '14 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

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

share|cite|improve this answer
I'm unfamiliar with the notations you used, do you mind letting me know what [N,G] means? Thanks! – Rod Apr 21 '14 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 '14 at 15:09
Why is [N,G] ≤ [N, G'] ? Shouldn't the inclusion go the other way? – TBrendle May 3 '15 at 23:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.