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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.

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    $\begingroup$ $((G,G),H)=1 \Rightarrow ((G,H),H)=1\wedge ((H,G),H)=1 \Rightarrow ((H,H),G)=1$. $\endgroup$ – Derek Holt Sep 29 '18 at 14:46
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    $\begingroup$ @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. $\endgroup$ – Bach Sep 29 '18 at 16:26
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    $\begingroup$ Yes and yes and the reasoning does work. $\endgroup$ – Derek Holt Sep 29 '18 at 16:37
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    $\begingroup$ 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. $\endgroup$ – C Monsour Sep 29 '18 at 17:13
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    $\begingroup$ 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! $\endgroup$ – Derek Holt Sep 30 '18 at 11:23
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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.

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  • $\begingroup$ There it is! Seems like commutators sometimes bring us tedious operations. Besides, I omitted the previous exercise bringing me to a deadlock. Thank you! $\endgroup$ – Bach Sep 30 '18 at 13:13
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    $\begingroup$ 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. $\endgroup$ – Derek Holt Sep 30 '18 at 13:58
  • $\begingroup$ @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 $ $\endgroup$ – Khoa ta Mar 29 '19 at 0:08
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    $\begingroup$ That follows from $[hk,g]=[k,g][h,g]$, which is proved in the first paragraph. $\endgroup$ – Derek Holt Mar 29 '19 at 2:30
  • $\begingroup$ @DerekHolt Thank you very much for pointing this out for me, your proof is very succinct compared to mine. $\endgroup$ – Khoa ta Mar 29 '19 at 3:00
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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 $$ = xb[b^{-1},x^{-1}]x^{-1}b^{-1} = 1 $$

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