I am working on this Exercise from Algebra by Hungerford (Exercise II.7.3(b)). It states
If $ H $ and $ K $ are subgroups of a group $ G $, let $ (H, K) $ be the subgroup of $ G $ generated by the elements $ \{ hkh^{-1}k^{-1}|h\in H, k\in K \} $. Show that
If $ (H, G')=\langle e \rangle $, then $ (H', G)=\langle e \rangle $.
$ G' $ is the commutator subgroup of $ G $.
My attempt: $ (H', G)= \langle e \rangle $ is the same thing as $ H' $ is in the center of $ G $. Then I am stuck… I couldn't find any useful tool to simplify the problem. Can someone give me a hint? Thank you.
Best Answer
By the previous exercise in the book, for $h,k \in H$, and $g \in G$, we have $[hk,g] = h[k,g]h^{-1}[h,g] = [k,g][h,g]$, because $(H,G')=1$ (I am writing $1$ for $\langle e \rangle$ and also for $e$.)
We have to prove that $[[h,k],g] = 1$. We have
$$[[h,k],g] = [hkh^{-1}k^{-1},g] = [k^{-1},g][h^{-1},g][k,g][h,g] = [k,g]^{-1}[h,g]^{-1}[k,g][h,g].$$
Now, using $(H,G')=1$ again, we have $h^{-1}[k,g]h=[k,g]$ and $hg^{-1}[k,g]gh^{-1} = g^{-1}[k,g]g$, and so $$[k,g]^{-1}[h,g]^{-1}[k,g][h,g] = [k,g]^{-1}ghg^{-1}h^{-1}[k,g]hgh^{-1}g^{-1}=[k,g]^{-1}[k,g]=1,$$ a required.