Let $H,K$ be two Abelian subgroups of a finite group $G$ such that $G=\langle H,K\rangle$. Show that $G'=[H,K]$.
My attempt: Ofcourse, $[H,K]\subseteq [G,G]=G'$. Conversely, let $x,y\in G$. It suffices to show that $[x,y]\in [H,K]$.
Since $G=\langle H,K\rangle$, there exists $a_i,b_i\in H\cup K$ such that $x=\prod_i a_i^{k_i}$ and $y=\prod_i b_i^{j_i}$. Thus $$[x,y]=x^{-1}y^{-1}xy=\prod_i a_i^{-k_i}\prod_i b_i^{-j_i}\prod_i a_i^{k_i}\prod_i b_i^{j_i}.$$
I don't know how to proceed from here. If $H$ and $K$ were to commute with each other, then I could proceed, but since they're only given to be Abelian, I can't seem to use this hypothesis.
I was able to solve using hints from @Cpc:
Let $x,y\in G$. Then there exists $u_i,w_i\in H\cup K$ such that $x=\prod_{i=1}^m u_i$ and $y=\prod_{i=1}^n w_i$. We proceed by induction on $m+n$.
Base cases:
$m+n=1$. Then $[x,y]=1\in [H,K]$.
$m+n=2$. Without loss of generality, we may assume $m=n=1$. If $u_1,w_1\in H$. Then $[x,y]\in [H,H]=1$, because $H$ is Abelian. Similarly, if $u_1,w_1\in K$, then again $[x,y]=1$. Next, if $u_1\in H, u_2\in K$, then $[x,y]=[u_1,w_1]\in [H,K]$. Lastly, if $u_1\in H, u_2\in K$, then $[x,y]=[u_1,w_1]\in [K,H]=[H,K]$.
Induction step: Suppose that $m+n\geq 3$. Without loss of generality, we may assume $m>1$. Then $$[x,y]=[xu_m^{-1}u_m,y]=[xu_m^{-1},y]^{u_m}[u_m,y]\in [H,K];$$using the induction hypothesis and $[H,K]\trianglelefteq G$.