Question

If K and N are normal subgroups of G such that ,prove that nk=kn for everyrole="math" localid="1652340816454" $n\in N, k\in K$ .

Short answer

It is proved that nk=kn for every $n\in N, k\in K$

Step-by-step solution

Referring to the result of Exercise 18

If N are K normal subgroups of G , then $K\cap N$ is a normal subgroup of G .

Proving that nk=kn for every n∈N,  k∈K

If N and K are a normal subgroup of G, then we know $K\cap N$ is also a subgroup of G (proved in exercise 18).

So, from the definition of the subgroup, we get:

$\left(nk\right)\in K\cap N$

${\left(nk\right)}^{-1}\in K\cap N$

Since it is given that

Which follows:

$\begin{array}{c}nk{\left(nk\right)}^{-1}=e\\ nk=kn\end{array}$

Therefore, it is proved that nk=kn for every $n\in N, k\in K.$