Is there a group homomorphism $f:G\longrightarrow G$ for which $G/\operatorname{Im} f \not\cong\operatorname{Ker} f$? $G$ is finite Can you find a counterexample to the claim that for all group homomorphisms $f:G \longrightarrow G$, $G/\operatorname{Im} f \cong \operatorname{Ker} f$.
Let $G = \mathbb{Z}$; $f(n) = 2n$ is a homomorphism. $\operatorname{Im} f = 2\mathbb{Z}$. $\operatorname{Ker} f = \{0\}$. $\mathbb{Z}/2\mathbb{Z} \not\cong \{0\}$.
What about an example for finite groups? 
What about for abelian finite groups?
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Here's an example, to justify Tobias Kildetoft's claim, of a homomorphism where the image is not normal. Let $D_6 = \langle x, y \mid x^3 = 1, y^2 = 1, yx = x^2y\rangle$ and $f:D_6 \longrightarrow D_6$ be such that $f(x) = 1$ and $f(y) = y$. $\operatorname{Im} f = \{1,y\}$, which isn't normal because $x(\operatorname{Im} f)x^{-1} \not=\operatorname{Im} f$.

For the abelian case consider $G=\mathbb Z/4\mathbb Z\oplus \mathbb Z/2\mathbb Z$, $f\colon (a+4\mathbb Z,b+2\mathbb Z)\mapsto (2b+4\mathbb Z,2\mathbb Z)$. Then $\ker f\cong \mathbb Z/4\mathbb Z$ and $G/\operatorname{Im} f\cong(\mathbb Z/2\mathbb Z)^2$.
