# 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.

• $((G,G),H)=1 \Rightarrow ((G,H),H)=1\wedge ((H,G),H)=1 \Rightarrow ((H,H),G)=1$. Commented Sep 29, 2018 at 14:46
• @DerekHolt Hi! Isn't $(G, H)=(G, H)$ by definition? And is $\wedge$ the symbol in logic meaning ''and''? If so, the reasoning might not work.
– Bach
Commented Sep 29, 2018 at 16:26
• Yes and yes and the reasoning does work. Commented Sep 29, 2018 at 16:37
• The last step in Derek's proof follows from the Three Subgroups Lemma. It's valid, but it's not a result someone posing this question would be familiar with. The statement and proof can be found on Wikipedia, and of course in many group theory texts, such as Isaacs. Commented Sep 29, 2018 at 17:13
• To me this exercise looks like an obvious application of the three subgroups lemma. But you are both right in saying that Hungerford has not covered that yet at this stage of the book. So I am unsure how he intended readers to solve this problem. The surrounding exercises are not particularly difficult. Either we are missing some more elementary solution, or the problem is more difficult than he thought - both ar epossible! Commented Sep 30, 2018 at 11:23

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.

• There it is! Seems like commutators sometimes bring us tedious operations. Besides, I omitted the previous exercise bringing me to a deadlock. Thank you!
– Bach
Commented Sep 30, 2018 at 13:13
• Once it is proved, the three subgroups lemma, which says $((H,K),L)=1 \wedge ((K,L),H)=1 \Rightarrow ((L,H),K)=1$ is very useful in solving problems like this while avoiding technicalities. Commented Sep 30, 2018 at 13:58
• @DerekHolt I don't know if I'm correct but in your proof, you use the fact that $[k^{-1}, g] = [k, g]^{-1}$ but I don't think this is correct since, $[k^{-1}, g ] [k, g] \neq 1$ Commented Mar 29, 2019 at 0:08
• That follows from $[hk,g]=[k,g][h,g]$, which is proved in the first paragraph. Commented Mar 29, 2019 at 2:30
• @DerekHolt Thank you very much for pointing this out for me, your proof is very succinct compared to mine. Commented Mar 29, 2019 at 3:00

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 $$= xb[b^{-1},x^{-1}]x^{-1}b^{-1} = 1$$

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 $$hlh^{-1}\color{red}{l^{-1}xl}\color{blue}hl^{-1}h^{-1}x^{-1}=hlh^{-1}(\color{red}g\color{blue}h\color{green}{g^{-1}h^{-1}})\color{green}{hg}l^{-1}h^{-1}x^{-1}=(ghg^{-1}h^{-1})hlh^{-1}hgl^{-1}h^{-1}x^{-1}=gh(g^{-1}lgl^{-1})h^{-1}x^{-1}=g(g^{-1}lgl^{-1})x^{-1}=lgl^{-1}x^{-1}=xx^{-1}=e$$