How to show if $H,K$ are normal subgroups in a group $G$ then $[H,K] \leq H \cap K$

How to show if $H,K$ are normal subgroups in a group $G$ then $[H,K] \leq H \cap K$.

Here $[H,K]$ is the commutator subgroup.

I have tried to start it but instantly failed:

I want to show firstly that $[H,K]\subseteq H \cap K$ but I don't understand what an arbitrary element of $[H,K]$ would even look like so I have something to work on.

I thought it could look something like $h^{-1}k^{-1}hk$ but then I thought that we are working with the smallest subgroup containing elements of this form but that might not be what the actual elements are?

Could anyone help?

Solution 1:

Recall that: $$[H,K] = \langle [h,k] = hkh^{-1}k^{-1}\ |\ h\in H,\ k\in K\rangle.$$ It suffices to show that the generators of $[H,K]$are contained in $H$ and $K$. From the above, a typical generator of $[H,K]$ looks like $hkh^{-1}k^{-1}$. Since $K$ is normal, you have $hkh^{-1} = k'$ for some $k'\in K$. Hence, $$hkh^{-1}k^{-1} = k'k^{-1}\in K.$$ Can you take it from here?