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Recently, one question was posed on embeddings of finite groups in $GL(n,\mathbb{Z})$ (here); it is mentioned that $S_n$ can be embedded in $GL(n-1,\mathbb{Z})$. How to prove it? Can we replace $\mathbb{Z}$ by finite field?

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1 Answer 1

$S_n$ naturally acts by permutations on a set of $n$ elements, which gives an $n$-dimensional permutation representation. This representation, like any permutation representation of any finite group, has a trivial subrepresentation. The quotient is then still defined over $\mathbb{Q}$ (and therefore also over $\mathbb{Z}$) and is $(n-1)$-dimensional, which gives the desired map to $\text{GL}_{n-1}(\mathbb{Z})$. In fact, in this particular case, the quotient is absolutely irreducible, since the permutation action is doubly transitive, but this is beside the point.

Since no non-trivial element of $S_n$ fixes all the elements in the set, the representation is faithful, thus the above map is an embedding. By composing with the reduction-modulo-$p$ map, you get a map to $\text{GL}_{n-1}(\mathbb{F}_p)$, and this is still an embedding.

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