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?

  • $\begingroup$ What does $C_G(x)$ denote? $\endgroup$ Oct 16, 2018 at 3:23
  • $\begingroup$ $N$ appear not to enter into your proposed conclusion. $\endgroup$ Oct 16, 2018 at 4:24
  • $\begingroup$ 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. $\endgroup$
    – kalikinik
    Oct 16, 2018 at 5:08
  • 1
    $\begingroup$ I am guessing that $x\in N$ is the assumption. $\endgroup$ Oct 16, 2018 at 8:26

2 Answers 2


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 $s\in C_G(x)$.

  • $\begingroup$ Wooow, good it is beatyful $\endgroup$ Oct 16, 2018 at 8:46

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


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..


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