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Aug 21 at 22:14 vote accept Fetchinson0234
Aug 21 at 20:13 history became hot network question
Aug 21 at 13:12 answer added Alexei Entin timeline score: 11
Aug 21 at 13:04 comment added Denis T @Smiley Infinite symmetric group has no finite dimensional representations by Jordan's theorem: every finite subgroup of GL(N) is abelian-by-(finite of bounded order); and because all Lie groups are abelian-by-linear, there's no such embedding.
Aug 21 at 13:00 answer added Denis T timeline score: 4
Aug 21 at 12:35 comment added Sean Eberhard The notion of limit of finitely generated groups was refined in the work of van den Dries and Wilkie. See sciencedirect.com/science/article/pii/0021869384902230. To apply this in the case of $S_N$ you would need to choose generators.
Aug 21 at 12:33 history edited Fetchinson0234 CC BY-SA 4.0
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Aug 21 at 12:32 comment added Sam Nead You might be interested in the concept of “Gromov-Hausdorff” convergence of (pointed) metric spaces. This applies to Cayley graphs, so can be used to find limits of sequences of groups, equipped with generating sets and a “rescaling” function.
Aug 21 at 12:25 comment added Smiley1000 I guess you can say that $\underset{\longrightarrow}{\operatorname{lim}} \mathbb{Z}_N \cong \mathbb{Q}/\mathbb{Z}$ embeds in $S^1$. In the same way, you can form the direct limit $\underset{\longrightarrow}{\operatorname{lim}} S_N$ which we might call $S_{(\infty)}$. I don't know whether this embeds in some Lie group.
Aug 21 at 12:13 history asked Fetchinson0234 CC BY-SA 4.0