If $H,K⊲G$ and $H∩K = \{1_G\}$, then all elements in $H$ commute with all elements in $K$ [duplicate]

Let $H,K⊲G$ be two normal subgroups such that $H∩K = \{1_G\}$. Prove all elements in $H$ commute with all elements in $K$.

I have no idea how to do this.

Solution 1:

In general, whenever you're given the condition that an intersection of two groups is trivial, you should consider trying to prove that some element is in each of the two groups, and hence it's the identity. In this case, you'd like to rephrase the question that "all elements in $H$ commutes with all elements of $K$" in a way that requires you to prove that some elements are the identity.

In this case, as Henning Makholm said, let $h\in H,k\in K$, then showing that $hk = kh$ is the same as showing that $hkh^{-1}k^{-1} = 1$. Hence, you'd like to show that the element $hkh^{-1}k^{-1} = 1$ for any $h\in H,k\in K$. But since you're given the fact that $H\cap K = \{1\}$, it suffices to prove that $hkh^{-1}k^{-1}$ is in both $H$ and $K$.

You should try to do this on your own. If you can't get it, here's the complete proof:

Note that $hkh^{-1}\in K$, since $K$ is normal, and hence $(hkh^{-1})k^{-1}$ is the product of two elements of $K$, hence it's also in $K$. Similarly, since $H$ is normal, $kh^{-1}k^{-1}\in H$, so $h(kh^{-1}k^{-1})$ is the product of two elements of $H$, hence it's also in $H$.

Hence, $hkh^{-1}k^{-1}\in H\cap K$, so $hkh^{-1}k^{-1} = 1$.