Let $\phi : G \rightarrow H$ be a homomorphism from group $G$ to group $H$. Let $g \in G$ and $h \in H$ such that $\phi(g)=h$…?

Let $\phi : G \rightarrow H$ be a homomorphism from group $G$ to group $H$. Let $g \in G$ and $h \in H$ such that $\phi(g)=h$. Let $A= \{g' \in G: \phi(g')=h\}$ and let $K$ be the kernel of the homomorphism $\phi$. Show that the set $A$ is equal to the left coset $gK$ of the subgroup $K$ of $G$.

I really don't know how to start this. I know I must transform $A$ into a form of $gK$ (right?), but don't know how to get there. Please help, I've been working on this problem for hours.

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One direction is easy to check: $gK\subseteq A$. Just apply $\varphi$ on the elements of $gK$.

Now, for the other, use that $a=g(g^{-1}a)$ for all $a\in A$, and that $$\varphi(g^{-1}a)=\varphi(g)^{-1}\varphi(a)=h^{-1}h=e,$$ from which your claim follows. Note that if $a=gf$ for some $f\in G$, then we necessarily have $f=g^{-1}a$. Hence the previous description.

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Hint:

let $\pi : G \to G/K$ be the standard projection, what is the image of $A$ under $\pi$?

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Erm, those sets are in fact equal, not just in bijection. Both aren't subgroups if $g\notin K$, but both are the same subset of $G$. Also the sets are $A$ and $gK$, not $A$ and $\{gK\}$! – Christoph Mar 3 '14 at 14:52
Yep, you're right. But the main idea holds, no? – Léo Mar 3 '14 at 22:57