# Prove that this is a surjection and find the kernel

We have a map $$\alpha: G \rightarrow S(G)$$, where $$S(G)$$ is the group of all bijections from $$G$$ to $$G$$. And $$\alpha(g) = f_g$$, where $$f_g(a) = gag^{-1}$$. It's easy to prove that this is a homomorphism and its kernel is the set of $$g\in G$$ such that $$f_g = \operatorname{Id}_G$$, i.e. $$f_g(a) = gag^{-1} = a$$ $$\forall a\in G$$. It means that $$ga = ag$$ $$\forall a\in G$$ and it is the definition of $$Z(G)$$, the center of group $$G$$.

But what about proving that this is a surjection?

I find it obvious by definition (I mean that $$\alpha(f_g) = g$$) But how to prove that this is a surjection strongly and what about my solution of kernel? Is it ok?

• It will not be a surjection in general. For a group of order $3$, $S (G)$ is a group of order $6$.
– cqfd
Mar 9 '19 at 13:02

In general, $$\alpha$$ is not surjective. This follows already from the fact that $$|S(G)|=n!$$ when $$|G|=n$$ (and $$n!>n$$ for $$n>2$$). Also, we have $$\alpha(g)(1)=1$$ for all $$g\in G$$, but (unless $$n=1$$) there exist bijections $$\in S(G)$$ that map $$1$$ elsewhere.
In summary, $$\alpha$$ is surjective iff $$G$$ is trivial.