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A definition of commutator subgroup of G is

$[G,G]:=\langle \left \{ [a,b] : a,b \in G \right \} \rangle$ (where $[a,b]=aba^{-1}b^{-1}$ ) I use the notation from this question from here, but use $[G,G]$ instead of $G'$. And the question is why the commutator subgroup of $G$ is a subgroup of $G$. Of course, according to the comment, the answerer recommended two alternative ways instead of the defintion of subgroup.In fact, I tried why $[G,G]$ become a subgroup of $G$ before I know two alternative methods and I am now confusing because I cannot assure my thought is right. To begin with, clearly I need to check

  • $[G,G]$ is closed by (indcued) operation $(G,*)$
  • The identity $e$ in $G$ is also in $[G,G]$
  • For any $x \in [G,G]$, there exists a (unique) inverse $x^{-1} \in [G,G]$ ....

To begin with, I think that

$x \in [G,G] \Longleftrightarrow x=g(aba^{-1}b^{-1}) $ or $x=g[a,b]$ for some $g \in G$ .... (★)

because $[G,G]$ is generated by $[a,b]$. It seems to be right when consiering the group of generated by sington element, but I cast doubt about my description when I checked each condition of the subgroup, especially the last condition.

$[G,G]$ is closed by (indcued) operation $(G,*)$

clearly, the operation (mulitplication on group $G$) is cleary well-operated in $[G,G]$, so for any $x,y \in [G,G]$, ($x=g_1[a,b], y=g_2[c,d]$, where $g_1,g_2 \in G $ and $a,b,c,d \in G$.)

$xy=g_1(aba^{-1}b^{-1})g_2[c,d]=(g_1aba^{-1}b^{-1}g_2)[c,d]$.

From here, since $g_1,g_2,b,a^{-1},b^{-1},g_2 \in G$, $g_1aba^{-1}b^{-1}g_2 \in G$. Hence, if my assumption is , (★) ,right, $xy$ is clearly in $[G,G]$.

The identity $e$ in $G$ is also in $[G,G]$ (I have no question about this)

For any $x \in [G,G]$, there exists a (unique) inverse $x^{-1} \in [G,G]$ ....

when any element $x=g_1[a,b]=g_1(aba^{-1}b^{-1}) \in [G,G]$ is given,

$e=g_1aba^{-1}b^{-1}bab^{-1}a^{-1}g_1^{-1}$ holds. Hence, I expect $bab^{-1}a^{-1}g_1^{-1}$ is an inverse element of $x$. The serious problem is I have not yet solved is whether or not this element is contained in $[G,G]$. i.e, $bab^{-1}a^{-1}g_1^{-1}=[b,a]g_1^{-1}\overset{??}{\in} [G,G]$ Clearly, $G$ is nonabliean case..(If not, $[a,b]=e$ and this is obviously nonsense ) So I cannot show that the element $[b,a]g_1^{-1}$ in $[G,G]$.

To sum up, my gist on the question is...

  1. is my statement (★) right?
  2. If (★) is right, how to check the last condition of the subgroup
Hanwoong Cho
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    No, $(\star)$ is not correct. Multiplying by $g\in G$ may get you out of the group generated (here by the commutators but also in general). Again, this depends on exactly what your working definition of "group generated by $X$". – user10354138 Sep 12 '21 at 05:49
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    The commutator subgroup consists of all possible (finite) products of commutators of elements in $G$. – Brauer Suzuki Sep 12 '21 at 06:42
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    The notation $\langle X\rangle$ literally means "the subgroup generated by the set $X$", that is, "the smallest subgroup of $G$ that contains $X$ as a subset". Since $[G,G]$ is defined as $\langle [x,y]\mid x,y\in G\rangle$, then it is a subgroup by definition. There is, literally, nothing to do to prove that it is a subgroup. – Arturo Magidin Sep 12 '21 at 07:13
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    @HanwoongCho: Define $[G,G]$ as the set of all products $x:=\prod_i [a_i,b_i]$, then it follows $[G,G]$ is closed under product. You find that $[a,b]^{-1}=[b,a]$, hence for any $x\in [G,G]$ it follows $x^{-1} \in [G,G]$ and it follows $[G,G] \subseteq G$ is a sub group. – hm2020 Sep 17 '21 at 08:48
  • I should have seen theorem 7.6 on Fraleigh textbook... , though, every answer is beneficial to me. :) – Hanwoong Cho Sep 17 '21 at 10:21

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Question: "To sum up, my gist on the question is...is my statement (★) right? If (★) is right, how to check the last condition of the subgroup?"

Answer: As mentioned in the comments above: Define for any $a,b\in G, [a,b]:=aba^{-1}b^{-1}$. It follows $[a,b]^{-1}=[b,a]$. If you define $[G,G]$ as the set of all finite products

$$x:=\prod_i [a_i,b_i]$$

for $a_i,b_i \in G$ it follows $[G,G]$ is closed under products and $e\in [G,G]$. For an element

$$x:=[a_1,b_1]\cdots [a_n,b_n]$$

it follows $x^{-1}:=[b_n,a_n][b_{n-1},a_{n-1}]\cdots [b_1,a_1] \in [G,G]$

and it follows $[G,G] \subseteq G$ is a subgroup.

hm2020
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