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?

  • $\begingroup$ It will not be a surjection in general. For a group of order $3$, $S (G) $ is a group of order $6$. $\endgroup$
    – 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.

  • $\begingroup$ Thank you so much! I have not thought about this obvious observation! $\endgroup$
    – ErlGrey
    Mar 9 '19 at 13:17

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