Suppose $ H\leqslant G $, prove that if $ (H, G')=\langle e \rangle $, then $ (H', G)=\langle e \rangle $. 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.
 A: 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.
A: Based on Dr. Derek Holt's idea, we want to show $ [a,b]x [a,b]^{-1}x^{-1} = 1 $ where $ a,b \in H, x \in G $.
The important fact that the hypothesis gives us is that $ (H,G') = 1 $ implies $ hg' = g'h$ for $ g' \in G', h \in H$.
Now, by prob 2 of Hungerford, we have that $ x[b,a]x^{-1} = [xb,a] [x, a]^{-1} $.  Then
$$  [a,b]x [a,b]^{-1}x^{-1} = [a,b]x [b,a] x^{-1} = [a,b] [xb,a] [x, a]^{-1} = [a,b] [xb,a] [a,x], $$
expand the first two terms out and simplifying, we have 
$$ = xbab^{-1}x^{-1}ba^{-1}b^{-1} [a,x] $$
Since $ b^{-1} \in H$, using the fact above, this gives
$$ = xbab^{-1}x^{-1}ba^{-1} [a,x] b^{-1} $$
Expand $[a,x]$ out and simplifying yields
$$= xbab^{-1}x^{-1}bxa^{-1}x^{-1}b^{-1} = xba[b^{-1},x^{-1}] a^{-1}x^{-1}b^{-1}$$
Using the fact above again and simplifying, we have
$$ = xb[b^{-1},x^{-1}]x^{-1}b^{-1} = 1 $$
A: The answer by @DerekHolt is great. I also tried to answer this question with the least computation and tedious work.
We aim to prove, for all $x,y \in G$ and $h, l \in H$, if $[x,y]$ commute with element of $H$, then $hlh^{-1}l^{-1}xlhl^{-1}h^{-1}x^{-1}=e$.
Consider $l^{-1}xl=g$, equivalently $x=lgl^{-1}$, then $$hlh^{-1}\color{red}{l^{-1}xl}\color{blue}hl^{-1}h^{-1}x^{-1}=hlh^{-1}(\color{red}g\color{blue}h\color{green}{g^{-1}h^{-1}})\color{green}{hg}l^{-1}h^{-1}x^{-1}=(ghg^{-1}h^{-1})hlh^{-1}hgl^{-1}h^{-1}x^{-1}=gh(g^{-1}lgl^{-1})h^{-1}x^{-1}=g(g^{-1}lgl^{-1})x^{-1}=lgl^{-1}x^{-1}=xx^{-1}=e$$
