Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 7 down vote accepted

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 $[g][h] = [gh]$)

share|cite|improve this answer
More generally, for any group $H$ (here $H=\Bbb Z$) the group $G$ acts on the space of functions $G\to H$ via $(g\cdot f)(x)=f(g^{-1}x)$, possibly with finite support imposed. By choosing $H$ to be finite, we get a faithful action of $G$ on a finite group. This is also the action used in the construction of wreath products. – blue Jul 10 '14 at 0:08

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.

share|cite|improve this answer

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.)


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.

share|cite|improve this answer
"Every group acts faithfully on itself by left multiplication." This is action of a group on itself as a set! Left multiplication map is not a group homomorphism. – Prism Jul 10 '14 at 2:03
Let $G$ be a group. Then we have left-multiplication action: for each $g\in G$, we have a map $\sigma_{g}: G\to G$ defined by $\sigma_{g}(x)=gx$. It is action of $G$ on the set $G$. Note that $\sigma_{g}$ is a set-theoretic bijection, but is not a group isomorphism. So the left-multiplication defines a map $\sigma: G\to S_{|G|}$ and not $\sigma: G\to\operatorname{Aut}(G)$. – Prism Jul 10 '14 at 2:04
After seeing your edit, I realize that the misunderstanding lies in the usage of the term "acting faithfully". I think asking for injective group homomorphism $G\to\operatorname{Aut}(H)$ is a more unambiguous question. In any case, thanks for taking your time to answer the question :) – Prism Jul 10 '14 at 2:14
Yes, and sorry. I was going off of the title and didn't quite think through your clarification in the highlighted portion of your question. At least the free group still works. – Andrew Kelley Jul 10 '14 at 2:19
No need to be sorry at all :) I do like your free group example! +1 – Prism Jul 10 '14 at 2:24

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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