Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
@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

1 Answer 1

up vote 3 down vote accepted

$S_n$ 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.

share|improve this answer
    
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
1  
I looked at it, and you are correct about what you need to verify. I recommend spending some more time trying to verify this yourself -- in the end, I think you will be enriched a lot by discovering the proof yourself. –  Adam Smith Oct 19 '11 at 2:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.