# Prove that $\varphi\big(C_G(x)\big) =C_H\big(\varphi(x)\big)$, where $\varphi:G\to H$ is a group homomorphism with certain properties.

Let $$N$$ be normal in $$G$$ and suppose that $$\varphi :G \to H$$ is surjective group homomorphism such that $$N \cap \ker(\varphi) =1$$. Show that $$\varphi\big(C_G(x)\big) =C_H\big(\varphi(x)\big)$$.

I am not sure how to start this problem. I know we will use the fact it's a surjective homomorphism, but will we use the fact that $$N$$ is normal?

• What does $C_G(x)$ denote? Oct 16, 2018 at 3:23
• $N$ appear not to enter into your proposed conclusion. Oct 16, 2018 at 4:24
• What exactly is $H$? Another arbitrary group? Subgroup? Or do you really want $N = H$? Once you clarify the notation, I would most definitely try a subset argument. Oct 16, 2018 at 5:08
• I am guessing that $x\in N$ is the assumption. Oct 16, 2018 at 8:26

I assume that $$C_\Gamma(t)$$ is the centralizer of an element $$t$$ in a group $$\Gamma$$. We have the following lemma.

Lemma. Let $$\phi:G_1\to G_2$$ be any homomorphism of groups $$G_1$$ and $$G_2$$. Then, $$\phi\big(C_{G_1}(g)\big) \subseteq C_{G_2}\big(\phi(g)\big)$$ for all $$g\in G_1$$.

Take an arbitrary element $$\gamma\in\phi\big(C_{G_1}(g)\big)$$. Then, $$\gamma=\phi(g')$$ for some $$g'\in C_{G_1}(g)$$. Thus, $$\gamma\,\phi(g)=\phi(g')\,\phi(g)=\phi(g'g)=\phi(gg')=\phi(g)\,\phi(g')=\phi(g)\,\gamma\,,$$ as $$g$$ and $$g'$$ commute. This means $$\gamma\in C_{G_2}\big(\phi(g)\big)$$.

I believe that the problem is the following. The OP probably missed the condition that $$x\in N$$. I shall delete or edit this answer if that is not the case.

Problem. Let $$N$$ be a normal subgroup of a group $$G$$. Suppose $$\varphi:G \to H$$ is surjective group homomorphism from $$G$$ to a group $$H$$ such that $$N \cap \ker(\varphi) =\text{Id}$$. Show that $$\varphi\big(C_G(x)\big) =C_H\big(\varphi(x)\big)$$ for all $$x\in N$$.

Therefore, the inclusion $$\varphi\big(C_G(x)\big)\subseteq C_{H}\big(\varphi(x)\big)$$ for all $$x\in G$$ is trivial by the lemma above. This inclusion is true even without the conditions that $$\varphi$$ is surjective, that $$N$$ is a normal subgroup of $$G$$, that $$x\in N$$, or that $$N\cap\ker(\varphi)=\text{Id}$$.

We shall now prove that $$C_{H}\big(\varphi(x)\big)\subseteq \varphi\big(C_G(x)\big)$$ for all $$x\in N$$. Fix an arbitrary element $$\eta\in C_{H}\big(\varphi(x)\big)$$. Since $$\varphi$$ is surjective, $$\eta=\varphi(s)$$ for some $$s\in G$$. We then have \begin{align}\varphi(x^{-1}s^{-1}xs)&=\big(\varphi(x)\big)^{-1}\,\big(\varphi(s)\big)^{-1}\,\varphi(x)\,\varphi(x)=\big(\varphi(x)\big)^{-1}\,\eta^{-1}\,\varphi(x)\,\eta\\&=\big(\varphi(x)\big)^{-1}\,\eta^{-1}\,\eta\,\varphi(x)=\big(\varphi(x)\big)^{-1}\,\varphi(x)=1_H\end{align} since $$\varphi(x)$$ commutes with $$\eta$$. Therefore, $$x^{-1}s^{-1}xs\in \ker(\varphi)$$.

Because $$x\in N$$, we have $$s^{-1}xs\in N$$ as $$N$$ is normal in $$G$$. Because $$N$$ is a subgroup of $$G$$, $$x^{-1}s^{-1}xs=x^{-1}\big(s^{-1}xs\big)\in N$$. In other words, $$x^{-1}s^{-1}xs\in N\cap\ker(\varphi)=\text{Id}\,.$$ Ergo, $$x^{-1}s^{-1}xs=1_G$$, whence $$xs=sx$$, and so $$s\in C_G(x)$$.

• Wooow, good it is beatyful Oct 16, 2018 at 8:46

Let $$\phi(y)\in \phi(C_G(x))$$ then

$$\phi(y)\phi(x)=\phi(yx)=\phi(xy)=\phi(x)\phi(y)$$

so $$\phi(y)\in C_H(\phi(x))$$ and $$\phi(C_G(x))\subseteq C_H(\phi(x))$$

Let $$y\in C_H(\phi(x))$$ than $$y=\phi(z)$$ because $$\phi$$ is surjective. You have that

$$\phi(zx)=\phi(z)\phi(x)=y\phi(x)=\phi(x)y=\phi(xz)$$

so $$h:=zx(xz)^{-1}\in \ker(\phi)$$.

By contraddiction if $$h\neq 1$$ then $$h\notin N$$ because $$N\cap \ker(\phi)=1$$...

Now I don’t understand what it could be..