Could someone explain why $S_{n}$ is isomorphic to a subgroup of $GL_{n}(\mathbb{R})$?
I've learned that groups with different sizes cannot be isomorphic, but also two groups that are the same size are not necessarily isomorphic. Also I know there is only 1 group with three elements, and there is only 1 group with two elements. So $S_{2}$ is isomorphic to $\mathbb{Z}_{2}$. But what about $S_{3}$ and $D_{3}$?
Thanks in advance.
Edit
In order to show that $S_{n}$ is isomorphic to a subgroup of $GL_{n}(\mathbb{R})$ which contains matrices with exactly one 1 in each row and column, I need to find a function $\phi$ such that $\phi(\sigma \tau) = \phi(\sigma)\phi(\tau)$ and show that $\phi$ is injective.
I'm a bit lost as to how to show the homomorphism and one-to-one property is satisfied.
Further Edit
Let $A, B$ be the permutation matrices you mentioned and $\sigma, \tau \in S_{n}$. Then $Ae_{j} = e_{\sigma(j)}$ and $Be_{j} = e_{\tau(j)}$. ($e_{j}$ is the standard basis vector)
I need to show: $\phi(\sigma \tau) = \phi(\sigma)\phi(\tau)$. If I let $A = \phi(\sigma)$ and $B = \phi(\tau)$, then I want to show $\phi(\sigma \tau) = AB$ right?
Best Answer
$S_n$ is isomorphic to a subgroup of $GL_n(\mathbb{R})$, namely the subgroup consisting of permutation matrices (matrices with exactly one $1$ in each row and column and zeros elsewhere).
EDIT : To prove the above fact, you should think about the meaning of the symmetric group. Elements of $S_n$ should permute the elements of the set $\{1,\ldots,n\}$. To figure out which permutation matrix corresponds to an element of $S_n$, you need to figure out how a permutation matrix permutes the elements of $\{1,\ldots,n\}$. Here's a hint : look at what a permutation matrix does to the coordinate vectors.
This will give you your map from $S_n$ to $GL_n(\mathbb{R})$; at that point; checking that it is a homomorphism should be easy.