Inverse image of a homomorphism

[tomkoolen]

The question: Let f: G -> H be a homomorphism of groups with ker(f) finite, the number of elements being n. Show that the inverse image is either empty or has exactly n elements.

My work so far:
Let h be eH (identity on H). Then the inverse image is ker(f) so has n elements, which makes it empty when n = 0.
I know how to do this for eH but how for all other y in H?

 

[nuuskur]

What makes you think any homomorphism is invertible?

EDIT: Scratch that, I might have mis-understood.
What is an inverse image, exactly? Do you mean the subset of H that is mapped from G?This is the kernel
Ker f := \{g\in G | f(g) = e\in H\}
assuming I understand the problem correctly
we also know that |Ker f| = n

The inverse image of just the unit element: f^{-1} (\{e\}) = \{e\in G, g_2,\ldots g_n\}\subset G

 

[Samy_A]

The question: Let f: G -> H be a homomorphism of groups with ker(f) finite, the number of elements being n. Show that the inverse image is either empty or has exactly n elements.

My work so far:
Let h be eH (identity on H). Then the inverse image is ker(f) so has n elements, which makes it empty when n = 0.
I know how to do this for eH but how for all other y in H?

I assume you meant that the inverse image of an element of H is either empty or has exactly n elements.

If $f(x)=f(y)$, what is $f(xy^{-1})$?
What does the answer tell you about the set $D=f^{-1}(h)$, where $h \in Im(f)$?