# Isomorphisms between Symmetric Groups, GL(R), and Dihedral Groups

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 $D_{3}$?

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?

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Since you've learned that groups with different sizes cannot be isomorphic, you could try to compare the sizes of $S_n$ and $GL_n(\mathbb R)$... –  Mariano Suárez-Alvarez Oct 19 '11 at 2:05
$D_3$ is the group of symmetries of a triangle, and the symmetries of a triangle are described by permutations of its vertices, so you get $S_3$. $S_n$ is not isomorphic to $GL_n(\mathbb R)$. The former is finite and the latter is infinite. –  Grumpy Parsnip Oct 19 '11 at 2:07
@Jim: Could you say the same thing about $S_{4}$ and $D_{24}$? –  Student Oct 19 '11 at 2:13
@Mariano: What about $S_{n}$ and a subgroup of $GL_{n}(\mathbb{R})$? –  Student Oct 19 '11 at 2:17
@Jon : $S_n$ 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). –  Adam Smith Oct 19 '11 at 2:19
$S_n$ 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.
I don't quite see how to show the map from $S_{n}$ to $GL_{n}(\mathbb{R})$ is a homomorphism. Could you look at my further edit? –  Student Oct 19 '11 at 2:53