If $f:G \to H$ is a homomorphism with kernel $N$ and $K$ is a subgroup of $G$ then prove that $f^{-1}(f(K))=KN$ If $f:G \to H$ is a homomorphism with kernel $N$ and $K$ is a subgroup of $G$ then prove that $f^{-1}(f(K))=KN$
Ok, so what I know from this: $G$ and $H$ are groups that must preserve operation. It is associative, has an identity and an inverse. If this is so, then I could possibly show that the identity of $G$ maps to the identity of $H$, therefore inverse if preserved? The concept of the kernel has always confounded me, not sure why. Any homomophism defines an equivalence relation. The kernel is that relation. If $K$ is a subgroup of $G$, then $K$ is nonempty and closed, has an identity and has an inverse. 
So, take $K < G$ Show $f^{-1}(f(K))=KN$ when ${kn/ K\in K, n \in N}$ and $ker f=N< G$
so we can say $f^{-1}(f(K)) \subseteq KN \subseteq f^{-1}(f(K))$
I know that there is a similar equation already on this site, but I still need help. Thanks
 A: This actually isn't too bad. We know that, by definition,
$$f^{-1}(f(K))=\{g\in G : f(g) = f(k)\text{ some }k\in K\}.$$
But if $f(g)=f(k)$ then

$$e_G=f(g)f(k)^{-1}=f(g)f(k^{-1})=f(gk^{-1})$$

This line uses the fact that $f$ is a homomorphism so that $f(a)^{-1}=f(a^{-1})$ and similarly $f(a)f(b)=f(ab)$.
From the definition of the kernel we have $gk^{-1}\in N$ i.e. $g\in kN\subseteq KN$, so $g\in KN$. Since $g$ can be any element of $f^{-1}(f(K))$, we have that $f^{-1}(f(K))\subseteq KN$.
Conversely:  if $kn\in KN$ we want to show that $kn\in f^{-1}(f(K))$ so we get the other inclusion. Let $kn\in KN$ then, we note--again by using the homomorphism property that $f(ab)=f(a)f(b)$ and the fact that $f(n)=e_H$ by the definition of the kernel--that

$$f^{-1}(f(kn))=f^{-1}(f(k)f(n))=f^{-1}(f(k)e_H)=f^{-1}(f(k))\in f^{-1}(f(K))$$

This shows that $KN\subseteq f^{-1}(f(K))$. Since $KN\subseteq f^{-1}(f(K))\subseteq KN$ we have that they are equal.
A: Let $x \in f^{-1}(f(K))$, then: $f(x) \in f(K)$, so $\exists$ $k \in K$ such that $f(x) = f(k)$. But:
$f(x) = f(k) \Rightarrow (f(k))^{-1}f(x) = e \Rightarrow f(k^{-1})f(x) = e \Rightarrow f(k^{-1}x) = e \Rightarrow k^{-1}x \in ker(f)$
$\Rightarrow k^{-1}x \in N \Rightarrow$ $\exists$ $n \in N$ such that $k^{-1}x = n$ $\Rightarrow x = kn$, some $n \in N$ and $k \in K $ $\Rightarrow x \in KN$.
The other direction is done just as above, reversing the process.
($e$ denotes the neutral element of $H$).
