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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?

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

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

user651267
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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