I have mostly a question in my part "2-" . But any correction will be cool.
Question:
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.
Q.E.D.
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 $g(xy)g^{-1} = (gxg^{-1})(gyg^{-1})$.