Prove that $G= \ker(f) \times{\rm im}(f)$

Question

Let $f:G\to G$ be a homomorphism such that $f\circ f = f$. Prove that if $im(f)$ is normal in $G$, then $G = \ker(f) \times{\rm im}(f)$. Hint: If $x\in G$, then $x = [xf(x^{-1})]f(x)$.

My first try:

We know ${\rm im}(f)\in G$. Let $g\in G$ but $g\not\in {\rm im}(f)$.
Let $x\in {\rm im}(f)$. Since ${\rm im}(f)$ is normal in $G$, we know: $$gxg^{-1} = x$$ Applying $f$ to both sides, we get: $f(gxg^{-1}) = f(x)$ and since $f$ is a homomorphism, we know: $$f(gxg^{-1}) = f(g)f(x)f(g^{-1})$$ So $f(g)f(x)f(g^{-1}) = f(x)$.

I need to some how prove that $f(g) = e$, so that $g\in \ker(f)$, does anyone know what to do next, or if this method will even work?

Answer 1

The hint shows that $\rm{im}f\cdot\rm{ker}f=G$, and follows from the homomorphism property.

It remains to show $\rm{im}f\cap\rm{ker}f=\{e\}$. Let $h\in\rm{im}f\cap\rm{ker}f$. Then $h=f(g)=f\circ f(g)=f(h)=e$.

Since both the image and the kernel of $f$ are normal, we have the result.

Answer 2

First: since $f\circ f= f$, then I claim that $\mathrm{Im}(f)\cap\ker(f)=\{e\}$. To see this, note that $\mathrm{Im}(f)\cap\ker(f)$ certainly contains $e$. For the converse inclusion, assume that $x\in\mathrm{Im}(f)\cap\ker(f)$. Then $f(x)=e$, and there exists $g\in G$ such that $f(g)=x$. But now, since $f\circ f=f$, applying $f$ again, we conclude... what?

Now, since $\mathrm{Im}(f)$ is normal by assumption, and $\ker(f)$ is normal always, and $\mathrm{Im}(f)\cap\mathrm{ker}(f)=\{e\}$, this tells us that $\langle \mathrm{Im}(f),\mathrm{ker}(f)\rangle=(\ker(f)) (\mathrm{Im}(f))\cong \mathrm{Im}(f)\times\mathrm{Im}(f)$.

So the only remaining thing is to show that every element of $G$ can be written as the product of something in the kernel and something in the image. That’s the point of the hint. Write $$x = x f(x)^{-1}f(x) = (xf(x^{-1}))f(x).$$ Clearly $f(x)\in\mathrm{Im}(f)$. So, we just need to show that $xf(x^{-1})\in \ker(f)$. To do that, we just need to apply $f$ and see if we get $e$. Do we?