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Now, the common introduction question to a commutator subgroup $G'$ is showing that it is normal in the group $G$. However, I'm having a problem with something even more basic than this.

Let $[a,b],[c,d] \in G'$,

Then we have:

$[a,b][c,d]=a^{-1}b^{-1}abc^{-1}d^{-1}cd$

I don't see an easy to conclude that this of the form $[g,h]=g^{-1}h^{-1}gh$ for $g,h \in G$.

What am i missing here?

$G' \neq \{ a^{-1}b^{-1}ab : a,b \in G \}$

$G' = \{ \langle a^{-1}b^{-1}ab \rangle : a,b \in G\}$

Shaun
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    How are you defining the commutator subgroup? The standard definition says it is the subgroup generated by commutators. Once can also define it to be any word in the elements of $G$ that has a rearrangement that multiplies to the identity, but that's not standard (I believe). – lulu Feb 01 '19 at 23:19
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    In general, the product of two commutator elements is not a commutator – Hagen von Eitzen Feb 01 '19 at 23:21
  • Oh. Then what is the operation that makes $G'$ a group? –  Feb 01 '19 at 23:22
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    It's the product inherited from $G$, of course. – lulu Feb 01 '19 at 23:22
  • But according to Hagen, this would mean $G'$ is not closed under the binary operation of G. –  Feb 01 '19 at 23:23
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    No. $G'$ is the subgroup generated by commutators, it is not the set of commutators (which, as you remark, is generally not closed under multiplication). – lulu Feb 01 '19 at 23:24
  • OOOOOOOH wow. Thanks! So I defind $G'$ wrong above. Okay –  Feb 01 '19 at 23:25

1 Answers1

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Note: $$\boxed{G'=\langle\{ [a, b]\mid a,b\in G\}\rangle.}$$ Hence for $[f,g], [h, k]\in G'$ with $f,g,h,k\in G$, we have that $[f,g][h,k]\in G'$, since it is a product of some generators of the subgroup.

More generally, for $\mathfrak{g}=\prod_{i\in I}[g_i, g'_i]$ and $\mathfrak{h}=\prod_{j\in J}[h_j, h'_j]$ belonging to $G'$, where $I$ and $J$ are some index sets for some element sequences $(g_i), (g'_i)\in G^I$ and $(h_j), (h'_j)\in G^J$, we have that

\begin{align} \mathfrak{g}\mathfrak{h}&=\prod_{i\in I}[g_i, g'_i]\prod_{j\in J}[h_j, h'_j] \tag{1} \\ &=\prod_{\ell\in L}[f_\ell, f'_\ell], \end{align}

where $L$ and the sequences $(f_\ell), (f'_\ell)\in G^L$ run through the commutators in $(1)$ in order; that is, one notices that it is a product of commutators as it is the product of two products of commutators.

As for showing it's a subgroup, I suggest the "two-step subgroup lemma". However, as pointed out in the comments below, it is much better to prove that the group $\langle X\rangle$ generated by a subset $X$ of a group $G$ with respect to the operation of $G$ is a subgroup of $G$ in general.

This question is about how to show the subgroup is normal.

Shaun
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    If it is defined as “the subgroup generated by...” then why would you need to show it is a subgroup? Did you mean “As for showing it’s a normal subgroup”? – Arturo Magidin Feb 02 '19 at 01:03
  • Well, for the sake of rigour, some of us prefer to check whether such definitions make sense. I think what I have typed in this answer is illustrative, at least, of how one works with $G'$ and what the general element of $G'$ is. I'll add a note giving a hint on how to prove that the subgroup is normal. – Shaun Feb 02 '19 at 10:38
  • D'you think that's a fair comment of mine, @ArturoMagidin? You seem more experienced than I am so your feedback would be appreciated $\ddot\smile$ – Shaun Feb 02 '19 at 10:54
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    There are two definitions of "the subgroup of group $G$ generated by subset $X$ of $G$". One is the intersection of all subgroups of $G$ containing $X$, and the other is the set of all products of elements of $X$ and their inverses. The first of these clearly does define a subgroup of $G$ and then you need to prove that the two definitions are equivalent. It is preferable to do all of that in a general setting, since then it becomes unnecessary to prove that you are defining a subgroup in specific cases. For normality, it is easier to prove the stronger property that $G'$ is characteristic. – Derek Holt Feb 02 '19 at 14:25
  • I've edited my answer, @DerekHolt. Thank you :) – Shaun Feb 02 '19 at 14:57
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    It would be very poor nomenclature if “the subgroup generated” created an object that was not in fact a subgroup... – Arturo Magidin Feb 02 '19 at 23:34
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    For general comments along the lines of Derek Holt’s comment, see this prior answer. – Arturo Magidin Feb 02 '19 at 23:36