Let $G$ be a group. Consider $G\to G, z\mapsto xzx^{-1}$. Is this injective? Is it surjective?

Question

Let $G$ be a not necessarily finite group. For $x \in G$, consider the map $G\to G, \, z\mapsto xzx^{-1}$. Is this map injective? Is this map surjective?

I believe that the map is injective because there exists an inverse map $y\mapsto x^{-1}yx$ that will return $z$.

However, I cannot figure out if it is surjective. My intuition says it is a bijection but I am not sure how to demonstrate that.

Answer 1

Yes, it is a bijection. And you have defined the inverse yourself: it is the map $z\mapsto x^{-1}zx$.

Answer 2

Fix any $y\in G$. Then $x^{-1}yx \mapsto x(x^{-1}yx)x^{-1}=y$. So the conjugation by $x$ is indeed surjective.