3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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
6
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you so much! I have not thought about this obvious observation! $\endgroup$
    – ErlGrey
    Mar 9 '19 at 13:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.