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…
- is my statement (★) right?
- If (★) is right, how to check the last condition of the subgroup
Best Answer
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.