I can see that if $H \unlhd \langle H, K \rangle$, then $[H,K] \le H$, but I don't see how can it be normal.
If $H$ is non-abelian and $H \unlhd \langle H, K \rangle$, how is $[H,K] \unlhd H$?
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group-theory
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1You need to prove that $[h_1,k]^{h_2} \in [H,K]$ and that follows from commutator identity number 3 in https://en.wikipedia.org/wiki/Commutator – 2017-01-09
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0Thanks so much. We get $[h_1, k]^{h_2} = [h_1 h_2, k][h_2, k]^{-1} \in [H, K]$. – 2017-01-09
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1More generally, we always have $[H,K] \unlhd \langle H,K \rangle$. – 2017-01-09