Showing commutator subgroup is a subgroup Now, the common introduction question to a commutator subgroup $G'$ is showing that it is normal in the group $G$. However, I'm having a problem with something even more basic than this.
Let $[a,b],[c,d] \in G'$, 
Then we have: 
$[a,b][c,d]=a^{-1}b^{-1}abc^{-1}d^{-1}cd$
I don't see an easy to conclude that this of the form $[g,h]=g^{-1}h^{-1}gh$ for $g,h \in G$. 
What am i missing here?
$G' \neq \{ a^{-1}b^{-1}ab : a,b \in G \}$
$G' = \{ \langle a^{-1}b^{-1}ab \rangle : a,b \in G\}$
 A: Note: $$\boxed{G'=\langle\{ [a, b]\mid a,b\in G\}\rangle.}$$ Hence for $[f,g], [h, k]\in G'$ with $f,g,h,k\in G$, we have that $[f,g][h,k]\in G'$, since it is a product of some generators of the subgroup.
More generally, for $\mathfrak{g}=\prod_{i\in I}[g_i, g'_i]$ and $\mathfrak{h}=\prod_{j\in J}[h_j, h'_j]$ belonging to $G'$, where $I$ and $J$ are some index sets for some element sequences $(g_i), (g'_i)\in G^I$ and $(h_j), (h'_j)\in G^J$, we have that
\begin{align}
\mathfrak{g}\mathfrak{h}&=\prod_{i\in I}[g_i, g'_i]\prod_{j\in J}[h_j, h'_j] \tag{1} \\
&=\prod_{\ell\in L}[f_\ell, f'_\ell],
\end{align}
where $L$ and the sequences $(f_\ell), (f'_\ell)\in G^L$ run through the commutators in $(1)$ in order; that is, one notices that it is a product of commutators as it is the product of two products of commutators.
As for showing it's a subgroup, I suggest the "two-step subgroup lemma". However, as pointed out in the comments below, it is much better to prove that the group $\langle X\rangle$ generated by a subset $X$ of a group $G$ with respect to the operation of $G$ is a subgroup of $G$ in general.

This question is about how to show the subgroup is normal.
