# Relationship between $S(G)$, $\text{Aut}(G)$

Let $$G$$ be a nontrivial group, denote $$\text{Aut}(G)$$ the group of all its automorphisms of $$G$$ and denote $$S(G)$$ the symmetric group on $$G$$, e.g. the set of all bijections $$f:G\rightarrow G$$. I would be interested, whether $$\text{Aut}(G)$$ and $$S(G)$$ can ever be isomorphic as a groups.

For a finite case, we can make the observation, that any permutation $$f\in S(G)$$ not fixing the identity cannot be automorphism, so in the finite case $$S(G)$$ and $$\text{Aut}(G)$$ aren't even equinumerous.

Could someone provide a rigorous argument for the infinite case? It seems like it is in fact impossible to have these two isomorphic as a groups, but can they atleast have the same cardinality?

We know that if $$|G|=\kappa$$ then $$|S(G)|=2^\kappa$$.

E: Adding some more ideas, someone could use:

Denote $$S_{fix}(G)$$ the set of permutations of $$G$$ that fix the identity element $$e\in G$$. Certainly we get $$\text{Aut}(G)\preceq S_{fix}(G)\preceq S(G)$$ If we can show that $$S_{fix}(G)$$ and $$S(G)$$ aren't equinumerous, then we are done proving that cardinalities cannot ever be equal.

So, the cardinality problem is solved, now can we show that $$\text{Aut}(G)$$ and $$S(G)$$ cannot be isomorphic in the infinite case (or construct a counterexample?).

• Note that $S(G)$ and $\operatorname{Aut}(G)$ are equinumerous if $|G|=1$. – Servaes Jul 14 '19 at 11:00
• Yeah, that's a trivial case, I added that I require $G$ to be nontrivial. – Michal Dvořák Jul 14 '19 at 11:01

Suppose $$C_\infty^\infty = \langle a_1 \rangle_\infty \times \langle a_2 \rangle_\infty \times \langle a_3 \rangle_\infty \times ...$$ is the direct product of countably many isomorphic copies of an infinite cyclic groups. Then $$|C_\infty^\infty| = \aleph_0$$. And we have $$S(\{a_1, a_2, a_3, ...\}) \leq Aut(C_\infty^\infty) \leq S(C_\infty^\infty)$$. Thus $$2^{\aleph_0} = |S(\{a_1, a_2, a_3, ...\})| \leq |Aut(C_\infty^\infty)| \leq |S(C_\infty^\infty)| = 2^{\aleph_0}$$, which results in $$|Aut(C_\infty^\infty)| = |S(C_\infty^\infty)|$$
• Could you please more elaborate why does $S(a_1,\ldots)$ inject into $Aut(C^\infty_\infty)$ ? – Michal Dvořák Jul 14 '19 at 18:15
• @MichalDvořák, this is a free abelian group of countable rank. Any permutation of its free generators induces an automorphism. You can understand elements of $C_\infty^\infty$ as finitary integer sequences (integer sequences with finitely many non-zero entries) equipped of an operation of entry-wise addition. If you change the order of the entries, you will: 1) still get finitely sequences, 2) this operation will "respect" addition, 3) this operation is invertible. Thus we can conclude, that it is an automorphism. – Yanior Weg Jul 14 '19 at 18:29