Commutator of a group

A commutator in a group $G$ is an element of the form $ghg^{-1}h^{-1}$ for some $g,h\in G$. Let $G$ be a group and $H\leq G$ a subgroup that contains every commutator.

$(a)$ Prove that $H$ is a normal subgroup of $G$.

$(b)$ Prove that the factor group $G/H$ is abelian.

I'm not sure what to do, any proofs or hnts are greatly appreciated. Thank you.

(a) $gxg^{-1}=\left(gxg^{-1}x^{-1}\right)x$

(b) $ghH=hg\left(g^{-1}h^{-1}ghH\right)=hgH$

First let's assume $H\trianglelefteq G$. To see $G/H$ is abelian is immediate, let $G'$ be the subgroup of $G$ generated by the commutators, then for any automorphism $\varphi$ and any $x\in G$, $\varphi(x)$ is a product of things of the form $\varphi(g)\varphi(h)\varphi(g)^{-1}\varphi(h)^{-1}$, hence $G'$ is characteristic so that $G/G'$ is a well-defined group. And because $\varphi(ghg^{-1}h^{-1})=1$ this means $\varphi(gh)=\varphi(hg)$, i.e. $G/G'$ is abelian.

But then there is a surjective homomorphism

$$G/G'\to G/H$$

given by reducing modulo $H$, and since the domain is abelian, so is the image. (this is just the third isomorphism theorem: $G/G'\bigg/ H/G'\cong G/H$)

Finally let's see that $H\trianglelefteq G$. Let $x\in G$, then

$$xHx^{-1}=\{xhx^{-1} : h\in H\}$$

since $xhxh^{-1}\in H$ we have that $xhx^{-1}\in H$, i.e. $xh=h'x$ for some $h'\in H$, i.e. $H\trianglelefteq G$ as desired.