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

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?

• Welcome; we use Mathjax for mathematical formulas, rather than ascii art of extended characters. Here’s a tutorial. Nov 30, 2020 at 2:21
• Note: the fact that $\mathrm{Im}(f)$ is normal means that if $x\in \mathrm{Im}(f)$ and $g\in G$, then $gxg^{-1}\in\mathrm{Im}(f)$. It does not mean that $gxg^{-1}=x$, as you assert in line 3. (I’m also assuming that $\mathrm{Im}(f)\in G$ is a typo, and you meant $\mathrm{Im}(f)\subseteq G$, or some other symbol). Given that you have no warrant to assert that $gxg^{-1}=x$, anything else you do after that is doomed to be incorrect. Nov 30, 2020 at 2:49

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?

• A couple typos on the last line, @Arturo. You clearly mean $\rm{ker}\color{red}{f}$ and that we get $\color{red}{e}$.
– user403337
Nov 30, 2020 at 3:05
• @ChrisCuster: Thanks. Nov 30, 2020 at 3:36

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.

• Oh yes, that was a mistake. @ArturoMagidin
– user403337
Nov 30, 2020 at 2:57