# Verification of a proof about relationship between normal subgroups and commutators

I would just like verification that the following implications are correct:

If $H$ and $K$ are normal subgroups of an arbitrary group $G$, then $H \cap K$ is a normal subgroup of $G$. But the fact that $H$ and $K$ are normal implies that $[H,K] \subset H \cap K$ (assume this was already proven - not a difficult proof). Then, (this is the key implication that I am questioning) this implies that $[H,K]$ is normal in G.

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Maybe the transitively helps $[H,K]$ here to be normal in the group; but this is very well-known result that:

If $G$ is not an abelian group, it is not necessarily true that: $$K\vartriangleleft H, H\vartriangleleft G\Longrightarrow K\vartriangleleft G$$

You can find through web some conditions which allow you keep transitively via normality and can find some counter examples as well.

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so how would I go about proving it without the transistivity? – afedder Oct 17 '12 at 8:30
@afedder: The way of Berci is enough for the conclusion. It is simple and so formal. I vote for him (+1). But for other problems, except this problem here, you should be cautious. :) – S. Snape Oct 17 '12 at 8:45

Yes, it seems so. $[H,K]$ (as the subgroup generated by all $[h,k]:=hkh^{-1}k^{-1}$ commutators) is indeed normal in $G$, because: $$g[h,k]g^{-1} = [ghg^{-1},gkg^{-1}] \in [H,K]$$

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but are my implications correct - that is my question...I already know that the end result is true – afedder Oct 17 '12 at 7:55