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Exhibit symmetric group as a subgroup of a general linear group

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.9

Solution: We know that if $F$ is a finite field then $\mathsf{Aut}(F^n) \cong GL_n(F)$. This isomorphism $\zeta$ can be defined as follows: given $\theta \in \mathsf{Aut}(F^n)$, $\zeta(\theta)$ is the matrix in $GL_n(F)$ whose $i$-th row is precisely $\theta(e_i)$. (In particular, $\zeta$ is not canonical since it depends on the choice and order of a basis; here we choose the standard basis.) In Exercises 5.1.7 and 5.1.8, we found an injective group homomorphism $\psi : S_n \rightarrow \mathsf{Aut}(F^n)$. Combining these results we have an injective group homomorphism $\Psi : S_n \rightarrow GL_n(F)$, computed as follows: $$\Psi(\pi) = [e_{\pi(1)}\ \cdots\ e_{\pi(n)}]^T.$$ We can see that each $\Psi(\pi)$ is obtained from the identity matrix by permuting the rows, so that each has a single 1 in each row and column and 0 in all other entries. Thus $S_n$ is identified with a subgroup of $GL_n(F)$ consisting of permutation matrices; moreover, by counting we see that all permutation matrices are represented this way.

Thus we have proven the result for a finite field $F$; we began with this case because it was shown previously that $\mathsf{Aut}(F^n) \cong GL_n(F)$. If this is true for arbitrary fields then the same proof carries over to arbitrary fields $F$; however this will not be proven until later in the text. In the meantime we can convince ourselves that the result holds over arbitrary fields $F$ by noting that, in computing the product of two permutation matrices, we never deal with numbers other than 1 or 0. In particular, the question of whether or not $k = 0$ in $F$ for some integer $k$ never arises, so that all computations hold for arbitrary fields (in which $1 \neq 0$) and in fact the set of permutation matrices over any $F$ is closed under matrix multiplication.


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