Show that $\phi(K)=KN/N$

Question

Let $N\triangleleft G$ and $K\leqslant G$. Consider $\phi :G \mapsto G/N$ onto group homomorphism. Show that $\phi(K)=KN/N$.

I thought using the equality $(G/N)^{n}=G^{n}N/N$ to show but I can not show. Any idea will be appreciated.

Answer 1

Suppose $gN \in \phi(K)$, then there exists $k \in K$ such that $gN = kN$ $\implies$ $k^{-1}g=n$ for some $n \in N$. So $kn=g$. Thus $gN = knN \in KN/N$. So $\phi(K) \subseteq KN/N$. Now suppose $x \in KN/N$ then $x=knN$ for some $k\in K, n \in N$ $\implies$ $x=knN=kN$ as $n \in N$ So $x \in \phi(K)$. So $ KN/N \subseteq \phi(K)$. Combining both we get our result.

Answer 2

You need to show it’s a subset and it covers everything. Both are straightforward ($knN=kN$ and any $knN$ has rep $kn$ is in the image of $k$. )

Clarification based on comments:$KN$ is the set of all elements $kn$ for $k$ in $K$ and $n$ in $N$. $KN/N$ is just the cosets of elements in $KN$. You don’t need that $N$ is normal for this to be true - that’s needed later to show nice things like that this is a group homomorphism. The key idea is that $\phi(k)=kN$ and $k=ke$ is in $KN$, so $\phi$ maps into $KN/N$. Then for the reverse, any closet is of the form $knN$, so it equals $\phi(k)$, so it’s in the image.