I have mostly a question in my part "2-" . But any correction will be cool.


Let $G$ be a group and we define the commutator as follow $[g;h]=ghg^{-1}h^{-1}$ with $g, h \in G$. We writte $D(G)$ the sub group of $G$ generated by the commutator. Prove that $D(G)$ is a normal subgroup of $G$.

My answer:

1- $D(G)$ is a sub group.
From the definition here we get that $ D(G) = [G;G] $ and so that any element of $D(G)$ is of the form $[g_1;h_1][g_2;h_2]...[g_n;h_n]$ for $n$ finite.
no emptyness: $e \in D(G)$ indeed $ \forall n \in \mathbb{N}$ with $g_i=e=h_i$ we have $[g_1;h_1][g_2;h_2]...[g_n;h_n]=e$
closure: If $h,g \in D(G)$ then $hg \in D(G)$ because $h$ and $g$ are both by definition a finite product of commutators hence $hg$ is too a final product of commutator.
inverse: it can be show (trivial) that $[g;h]^{-1}=[h;g]$ but on an other side we have that $[g;h][h;g]=e$ so from here it is easy to show by generalization that $\forall x \in D(G)$ his inverse $x^{-1}$ can be express as a finite multiplication of commutator.

2-$D(G)$ is a normal sub group of $G$
We need to show that $\forall g \in G, x \in D(G) \Rightarrow gxg^{-1} \in D(G)$.
By definition $x=[g_1;h_1][g_2;h_2]...[g_n;h_n]$ so $gxg^{-1} =g [g_1;h_1][g_2;h_2]...[g_n;h_n] g^{-1} $
To simplify lets focus on $g[g_i;h_i]g^{-1}$ and prove that this is too a commutator.
$g[g_i;h_i]g^{-1}= gg_ih_ig_i^{-1}h_i^{-1}g^{-1}$. Now we rearange and we get that this is equivalent to $[gg_ig^{-1}; gh_ig^{-1}]$ which is itself a commutator in $G$.

I have mostly a question in my part "2-" is it correct? but any correction will be cool.


  • 2
    $\begingroup$ Please read the solution-verification tag description : "For posts looking for feedback or verification of a proposed solution. "Is my proof correct?" is off topic (too broad, missing context). Instead, the question must identify precisely which step in the proof that is in doubt, and why so. " $\endgroup$
    – Digitallis
    Dec 26, 2023 at 19:02
  • $\begingroup$ @Digitallis I am going to improve my post when i lk be able $\endgroup$
    – OffHakhol
    Dec 26, 2023 at 19:08
  • 2
    $\begingroup$ A very minor nit: I usually see the commutator of two elements denoted as $[g, h]$ (using a comma, not a semicolon) and the commutator subgroup denoted as $[G, G]$. $\endgroup$ Dec 26, 2023 at 19:13
  • $\begingroup$ math.stackexchange.com/questions/1786080/… $\endgroup$
    – Fuat Ray
    Dec 27, 2023 at 18:20

1 Answer 1


You don't need to prove anything specific to commutators for part 1: by definition if $X$ is a subset of the elements of a group $G$, by definition, the subgroup of $G$ generated by $X$ is the smallest subgroup of $G$ containing $X$. It is given by the intersection of all subgroups of $G$ that contain $X$ and you can verify that this is a subgroup and is contained in any subgroup that contains $X$.

For part 2, you need to justify why it is sufficient to show that for each $g \in G$, the set of commutators is closed under the mapping $x \mapsto gxg^{-1}$ (it is not enough just to say "to simplify": you need to justify the simplification). The key point is that this mapping (known as conjugation by $g$) is a group homomorphism: for any $x, y$, we have $g(xy)g^{-1} = (gxg^{-1})(gyg^{-1})$.

  • $\begingroup$ Thk a lot for your answer i ll read it when i can! $\endgroup$
    – OffHakhol
    Dec 26, 2023 at 19:17
  • $\begingroup$ If you want to avoid your question being closed, I suggest you mark the parts of your proof that you are unsure about. $\endgroup$
    – Rob Arthan
    Dec 26, 2023 at 20:01
  • $\begingroup$ the part "2-" now i ven't a n access to a computer so I can't do it. $\endgroup$
    – OffHakhol
    Dec 26, 2023 at 20:04
  • $\begingroup$ Thk for your advice $\endgroup$
    – OffHakhol
    Dec 26, 2023 at 20:04

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