I am looking left and right for a lemma to solve a problem on solvable group here, and I think I have found one under commutator group:
For any two subgroups $H, K$ of $G$, the $[H, K]$ is a normal subgroup of $\langle H, K \rangle$.
To be sure I know the angle brackets "$\langle$" and "$\rangle$" stand for generator, and the commutator of $H$ and $K$ stands for:
$$\begin{align} [H, K] :&= \langle [h, k] \mid h \in H, k \in K \rangle \\ &= \langle h^{-1}k^{-1}hk \rangle , \end{align}$$
but I don't know (or remember) and definition of $\langle H, K \rangle$. I would love to see the set builder of this $\langle H, K \rangle$, thank you for your time.
$\langle H, K \rangle$ is the minimum subgroup of $G$ under inclusion containing both $H$ and $K$, it is also the intersection of all the the subgroups containing $H$ and $K$. basically it is $\langle H \cup K \rangle$
You can also look at it as the set of all finite products of the form $s_1s_2\dots s_n$ where $s_i\in H \cup K$. When talking about an arbitrary set $S$ we let $s_i$ or $s_i^{-1}\in S$, but since $H$ and $K$ are subgroups we can dispense of this restriction.
Proof that $H'=[H,H]$ is a normal subgroup of $H$:
it suffices to show that if we take an element $h'$ of $H'$ and an arbitrary element $x$ of $H$ then $xh'x^{-1}=h''$ for some $h''$ in $H$.
This is the same as saying that $xh'=h''x$ for some $h''$ in $H'$. Here is when the commutator comes in. Notice that $[a,b]ba=aba^{-1}b^{-1}ba=ab$ (This is why it is called the commutator)
Therefore $[x,h']h'x=xh'x^{-1}h'^{-1}h'x=xh'$. So the $h''$ we are looking for is $[x,h]h'$ which clearly does belong to $H'$