Does every group act faithfully on some group?

Cayley's Theorem shows that every group acts faithfully on some set. In other words, one can find an injective group homomorphism $\sigma: G\to S_{A}$ where $S$ is the set of all bijections on some set $A$.

Can we prove that every group acts faithfully on some group? In other words, given a group $G$, does there exist an injective group homomorphism $\sigma: G\to\operatorname{Aut}(H)$ for some group $H$. Here $\operatorname{Aut}(H)$ is the group of all (group) isomorphism on $H$.

Partial answer: When $G$ has trivial centre, consider the conjugation map $\sigma: G\to\operatorname{Aut}(G)$ defined by $\sigma(g): G\to G$ where $\sigma(g) (a) = g a g^{-1}$. It is easy to see that $\sigma$ is injective.

What about when $G$ does not have a trivial centre?

Solution 1:

The first example that comes to mind is $\mathbb{Z}[G]$, the free abelian group generated by the symbols $[g]$ for $g \in G$.

The action of $G$ on $\mathbb{Z}[G]$ is induced by $g[h] = [gh]$ and is a faithful one.

(in fact, we can give $\mathbb{Z}[G]$ a ring structure, by defining $[g][h] = [gh]$)

Solution 2:

I think I have found an answer to my own question but not too sure:

Use Cayley's Theorem to find an injective group homomorphism $\sigma: G\to S_{A}$ where $S_{A}$ is symmetric group on some set $A$. We can assume $|A|\geq 3$. Since $S_{A}$ has a trivial centre (proof here), the conjugation action gives an injective group homomorphism $\phi: S_{A}\to\operatorname{Aut}(S_{A})$.

The composition $\phi\circ\sigma: G\to\operatorname{Aut}(S_{A})$ is an injective group homomorphism.

Solution 3:

Yes, $F_G$, the free (non-abelian) group, would work as the group on which $G$ acts. (It's automorphism group contains the symmetric group $S_G$ as a subgroup, and any such group would work.)

Edit:

Every group acts faithfully on itself by left multiplication. But as the OP's comment below shows, that isn't what his highlighted question asks.

In reference to plain old (set) actions, notice that the axioms of group actions assume no structure at all on the set. So if $G$ acts on a set $X$ faithfully, simply put any group structure on $X$, and we now have $G$ acting faithfully on a group.

Solution 4:

Your idea about using groups with non-trivial center can be extended: Simply embed your given group $G$ into a group with trivial center, and use the same map which you give. For example, take the free product of $G$ with the infinite cyclic group $H=G\ast\mathbb{Z}$, then $G$ acts faithfully by $\gamma_g:w\mapsto gwg^{-1}$.

Note that this idea retains certain finiteness properties of the group $G$, as the group $H$ is finitely presentable (finitely generated) if $G$ is finitely presentable (finitely generated). In the other answers, the groups $F_G$ and $S_G$ are finitely generated if and only if $G$ is finite.