# Bijection on conjugate equivalence induce isomorphism?

Given group $$G$$, we denote $$G'=G/$$~, where "~" is conjugate equivalence ($$a$$~$$b$$ iff there exists $$c$$, s.t. $$a=c^{-1}bc$$) . Then, a group homomorphism $$f:G\to H$$ induces a map in sets of conjugate equivalent classes $$f':G'\to H'$$.

Now, suppose the induced map $$f'$$ is a bijection, is it true that the original group homomorphism $$f$$ is an isomorphism?

I think the hard part is proving $$f$$ is surjective. And in finite case, I tried to use Class Equation of Group to work it out. But I still got stuck and couldn't go any further. Hope someone could help. Thanks!

If $$H$$ is a finite group, the answer is YES (I don't know how to conclude if $$H$$ is infinite).

If $$g\in G$$, let use denote by $$\gamma_G(g)$$ its equivalence class, i.e. the set $$\gamma_G(g)=\{ c^{-1}g c\mid c\in G\}$$, and let's use a similar notation for $$H$$.

Let $$f:G\to H$$ be a group morphism such that the map $$f':\gamma_G(g)\in G'\mapsto \gamma_H(f(g))\in H'$$ is bijective.

First step. $$f$$ is injective (it works even if $$H$$ is not supposed to be finite).

For, let $$g\in G$$ such that $$f(g)=1_H$$. Then $$f'(\gamma_G(g))=\gamma_H(f(g))=\gamma_H(1_H)=f'(\gamma_G(1_G))$$, and by assumption, $$\gamma_G(g)=\gamma_G(1_G))=\{1_G\}$$. Thus , $$g$$ is conjugate to $$1_G$$, which easily implies that $$g=1_G$$.

Second step. Let $$K=im(f)$$. Then $$H=\displaystyle\bigcup_{h\in H}hKh^{-1}$$.

Indeed, let $$h\in H$$. By assumption, there exists $$g\in G$$ such that $$\gamma_H(h)=f'(\gamma_G(g))=\gamma_H(f(g))$$. It means that $$h$$ is conjugate to $$f(g)\in K$$, hence the desired equality.

Third step. Assume that $$H$$ is finite (and thus so is $$G$$, since $$f$$ is injective). Then $$K=H$$ , that is , $$f$$ is surjective.

Indeed, it is a classical fact that a finite group cannot be the union of conjugates of a proper subgroup.

Let's give a quick proof of this fact. Let $$U= \displaystyle\bigcup_{h\in H}hKh^{-1}$$. Let $$h_1,\ldots,h_r$$ be the representatives of the left cosets of $$H$$ modulo $$K$$. Then it is easy to see that $$U\setminus\{1_H\}=\displaystyle\bigcup_{i=1}^r (h_iKh_i^{-1}\setminus\{1_H\})$$.

In particular, $$\vert U\vert-1\leq r(\vert K\vert -1)=\vert H\vert-[H:K]$$. If $$U=H$$, then $$[H:K]\leq 1$$, so $$[H:K]=1$$ and $$K=H$$.

Note that $$GL_2(\mathbb{C})$$ is the union of the conjugates of the subgroup of upper triangular matrices, so the third step does not hold in general in the infinite case.

• Your notation for conjugacy classes is very confusing. It conflicts with the standard notation of centralizers. I would write $g'$ in this context for consistency. – Orat Oct 9 at 1:51