2

I know this is a duplicate of Suppose $ H\leqslant G $, prove that if $ (H, G')=\langle e \rangle $, then $ (H', G)=\langle e \rangle $. but the only answer which does not use the Three Subgroups Lemma (which, at this point, Hungerford has not covered) did not make sense to me, and since my reputation is still low, I can't leave a comment asking for clarification. The given solution is as follows:

"By the previous exercise in the book, for $h,k∈H$, and $g∈G$, we have $[hk,g]=h[k,g]h^{−1}[h,g]=[k,g][h,g],$ because $(H′,G)=1$ (I am writing 1 for ⟨e⟩ 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

$$[k,g]^{−1}[h,g]^{−1}[k,g][h,g]=[k,g]^{−1}ghg^{−1}h^{−1}[k,g]hgh^{−1}g^{−1}=[k,g]^{−1}[k,g]=1$$ as required."

The problem I have with this proof is that they seem to be using $(H',G)=1$, which is what we're trying to prove; we have that $(H,G')=1$ and we're trying to prove $(H',G)=1$, so it seems to me we shouldn't assume that in our proof. Am I missing something? If not, and this proof is invalid, would anyone be able to provide a valid proof of this (without referring to the Three Subgroups Lemma)?

Joe
  • 105
  • 1
    Sorry, the proof was mine, and when I wrote "using $(H',G)=1$", I meant "using $(H,G')=1$", which we are assuming. So it was just a typo, both times. I have edited the solution now to remove the mistake. – Derek Holt Nov 18 '18 at 11:59
  • Gotcha! Thanks for the edit, I am able to follow most of the proof now. I do have one more question, though; How did you get that $[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g]$? I wrote it out and it doesn't seem to be equal at first glance; were you using something to get this equivalence? – Joe Nov 18 '18 at 19:30
  • $[k^{-1},g]=[k,g]^{-1}$ follows from $[hk,g]=[k,g][h,g]$, which we have just proved. – Derek Holt Nov 18 '18 at 22:18
  • @DerekHolt: Since the community has not voted to close this question, would you be willing to take these two comments and put them into an answer so that we can at least remove it from the Unanswered list? – aleph_two Dec 20 '18 at 05:43

1 Answers1

1

The answer to the linked question referred to in this query was by me. There was a mistake (or perhaps typo!) in that answer, which I have now corrected. When I wrote "Using $(H',G)=1$", I meant "using $(H,G')=1$.

For the second query in a comment, $[k^{-1},g]=[k,g]^{-1}$ follows from $[hk,g]=[k,g][h,g]$, which had been proved at that point.

Derek Holt
  • 90,008