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?).

  • 1
    $\begingroup$ Note that $S(G)$ and $\operatorname{Aut}(G)$ are equinumerous if $|G|=1$. $\endgroup$
    – Servaes
    Jul 14, 2019 at 11:00
  • $\begingroup$ Yeah, that's a trivial case, I added that I require $G$ to be nontrivial. $\endgroup$ Jul 14, 2019 at 11:01

1 Answer 1


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 $|Aut(C_\infty^\infty)| = |S(C_\infty^\infty)|$

  • $\begingroup$ Could you please more elaborate why does $S(a_1,\ldots)$ inject into $Aut(C^\infty_\infty)$ ? $\endgroup$ Jul 14, 2019 at 18:15
  • $\begingroup$ @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. $\endgroup$ Jul 14, 2019 at 18:29

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