Suppose $N \cap K = \{e_G\}$. Prove that for all $a \in N$ and $b \in K, ab = ba.$

Question

Let $G$ be a group and let $N$ and $K$ be normal subgroups of $G.$ Suppose $N \cap K = \{e_G\}$. Prove that for all $a \in N$ and $b \in K, ab = ba.$

I am seeing that all normal subgroups are not necessarily abelian so I am not sure what tools to use for this one.

Answer 1

By normality, $aba^{-1}\in K$. And by normality of $N$, $ba^{-1}b^{-1}\in N$.

Now $(aba^{-1})b^{-1}\in K$ and $a(ba^{-1}b^{-1})\in N$.

Thus $aba^{-1}b^{-1}\in N\cap K=\{e\}$.

We can conclude that $ab=ba$.

Answer 2

Note that $ab(ba)^{-1}$ is in both $N$ and $K$ by definition. Hence, $ab(ba)^{-1}$ must be equal to $1_G.$ This proves the result. Lemme know if you have any question.

A little more elaboration:

$ab(ba)^{-1}= (aba^{-1})b^{-1}.$ Note that $aba^{-1}$ and $b^{-1}$ both are in $K.$ Therefore, their product has to be in $K.$ Similar argument shows that $ab(ba)^{-1}$ is in $N.$