Finitary symmetric group is not injective endomorphism-invariant in symmetric group

Statement

Suppose $S$ is an infinite set. Denote by $G=\operatorname {Sym} (S)$ the symmetric group on $S$ and by $H=\operatorname {FSym} (S)$ the finitary symmetric group on $S$, viewed as a subgroup of $G$.

Then, $H$ is not an injective endomorphism-invariant subgroup of $G$. In other words, there exists an injective endomorphism $\sigma$ of $G$ such that $\sigma (H)$ is not contained in $H$.

Related facts

• Finitary symmetric group is characteristic in symmetric group

Axiomatic assumptions

For certain cardinalities of $S$ (including the countably infinite cardinality), the proof does not rely on the axiom of choice. However, the assertion for all infinite $S$ does rely on the axiom of choice.

Proof

We construct a bijection $\alpha$ between $S\times \mathbb {N}$ and $S$. The bijection is explicit for some infinite cardinalities of $S$, and its existence is guaranteed by infinite cardinal arithmetic for all infinite $S$ by the axiom of choice. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.

We now consider the composite mapping:

$\operatorname {Sym} (S)\to \operatorname {Sym} (S\times \mathbb {N} )\to \operatorname {Sym} (S)$

The first homomorphism simply operates by acting only on the first of two coordinates, ignoring the second coordinate. The second homomorphism operates using the bijection $\alpha$. Let $\sigma$ be the composite mapping. The following are easy to check:

• $\sigma$ is injective: The first map in the composite is injective, and the second map is bijective, so the composite is injective.
• $\sigma$ does not preserve $H$: Any element of $H$ with nontrivial support gets sent to something that moves infinitely many elements.