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By Cayley's theorem, any group $G$ naturally injects into the symmetric group $\mathrm{Sym}(G)$ of its underlying set via the Cayley embedding. Let's say this embedding is optimal for $G$, if there is no embeding $G \hookrightarrow \mathrm{Sym}(I)$ for some smaller index set $I$.

The finite groups with this property are completely understood, they are just the primary cyclic groups, the generalized quaternion groups, or $C_2\times C_2$.

For infinite groups, this phenomenon is a lot more common: In fact, if $|G|$ is a strong limit cardinal (that is, if $\lambda<|G|$ implies $2^\lambda<|G|$), it can have no smaller faithful permutation representation (as $|\mathrm{Sym}(\lambda)|=2^\lambda$ for any infinite $\lambda$).

But if $|G|$ is not a strong limit, I do not know what happens in general. Are there any groups of these sizes with no smaller embeddings? Is there even a characterization of them?

Some thoughts: With a bit of work, I was able to show that any infinite abelian group $A$ with $|A|$ not a strong limit always has a more efficient embedding. I will reproduce the proof in case it could lead to any insights into the general case:

Suppose $A$ is such a group. Fix some $\lambda<|A|$ for which we have $|A|\leq 2^\lambda$. We will show that $A$ then embeds into $\mathrm{Sym}(\lambda)$. To that end, we will show that there is an embedding $\iota: A \hookrightarrow \left(\mathbb{Q}/\mathbb{Z}\times \mathbb{Q}\right)^\lambda$. This will suffice: The subgroups $$A_\alpha= \ker(\pi_\alpha\circ \iota) $$ (where $\pi_\alpha$ is the projection onto the $\alpha$-th coordinate) are each of (at most) countable index in $A$ and altogether intersect trivially (both clear by definition). Hence, the action of $A$ on the disjoint union $\coprod_{\alpha<\lambda}A/A_\alpha$ is faithful on a set of cardinality $\lambda\times \omega=\lambda$.

It thus remains to construct this embedding $\iota$. First, we deal with the $\mathbb{Q}/\mathbb{Z}$-part: Fix any prime $p$ and consider the $\mathbb{F}_p$-vector space $V_p$ of the $p$-torsion elements. Since $|V_p|\leq |A|\leq 2^\lambda$, we have an embedding $V_p \hookrightarrow \mathbb{F}_p^\lambda$ which further embeds into $\left(\mathbb{Q}/\mathbb{Z}\right)^\lambda$ (just coordinatewise). Since this latter group is injective, this embedding extends to a homomorphism $\iota_p:A\to \left(\mathbb{Q}/\mathbb{Z}\right)^\lambda$ which is then injective on the $p$-component of $A$. If we put all these homomorphisms for each prime $p$ together, we obtain a map $$\iota_T:A\to \left(\mathbb{Q}/\mathbb{Z}\right)^{\lambda\times \omega}\simeq \left( \mathbb{Q}/\mathbb{Z} \right)^{\lambda}$$ that is injective on the torsion subgroup $T$ of $A$. Finally, the torsion free quotient $A/T$ likewise embeds into $\mathbb{Q}^\lambda$, which we can then amalgamate with $\iota_T$ into a map into $\left(\mathbb{Q}/\mathbb{Z}\times \mathbb{Q} \right)^\lambda$ which is then finally guaranteed to be injective.

Tim Seifert
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    I would imagine a Jonsson group of size $\omega_1$ (that is, a group of size $\omega_1$ such that every proper subgroup is countable) does not embed in $\mathrm{Sym}(\omega)$, but I've thought about it a few days now and can't find a proof. – Brian Pinsky Feb 26 '24 at 04:28
  • Sounds good! Any such group must have an uncountable orbit under any nontrivial action and as such cannot act faithfully on $\omega$. – Tim Seifert Feb 26 '24 at 09:44
  • Oh yeah that really should have been obvious. – Brian Pinsky Feb 26 '24 at 13:17

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