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It is well known that any finite group can be embedded in Symmetric group $S_n$, $GL(n,q)$ ($q=p^m$) for some $m,n,q\in \mathbb{N}$. Can we embed any finite group in $A_n$, or $SL(n,q)$ for some $n,q\in \mathbb{N}$?

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you can also embed $G$ in $S_n$ then embed $S_n$ in $SL$ as $\{0,1\}$-permutation matrices. –  yoyo Mar 15 '11 at 15:24

3 Answers 3

up vote 7 down vote accepted

Yes.

The symmetric group Sym(n) is generated by { (1,2), (2,3), …, (n−1,n) }. You can embed Sym(n) into Alt(n+2) as the group generated by { (1,2)(n+1,n+2), (2,3)(n+1,n+2), …, (n−1,n)(n+1,n+2) }. This embedding takes a permution π in Sym(n) and sends it to π⋅(n+1,n+2)sgn(π), where sgn(π)∈{0,1} is the parity of the permutation.

In other words, G ≤ Sym(n) ≤ Alt(n+2) embeds any group into a (slightly larger) alternating group.

The general linear group GL(n,q) embeds in the special linear group SL(n+1,q) using a determinant trick. We just add a new coordinate to cancel out the determinant of the matrix from GL(n,q) so the result lands in SL(n+1,q).

$$\operatorname{GL}(n,q) \cong \left\{ \begin{bmatrix} A & 0 \\ 0 & 1/\det(A) \end{bmatrix} : A \in \operatorname{GL}(n,q) \right\} ≤ \operatorname{SL}(n+1,q)$$

In other words, G ≤ GL(n,q) ≤ SL(n+1,q) embeds any group into a (slightly larger) special linear group.

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Hi Jack, I have a question, I'm a little bit confused with the map $\pi \mapsto \pi \cdot (n+1, n+2)^{\color{red}{\text{sgn}(\pi)}}$, what do you mean by that 'exponent'? Isn't it true that $(n+1, n-2)^ {-1} = (n+1, n+2)^1 = (n+1, n+2)$? Am I missing something here? Thanks a lot. :) –  user49685 Dec 11 '13 at 11:20
    
@user49685: Looks like sgn(π) should be 0 or 1 instead. –  Jack Schmidt Dec 11 '13 at 15:47
    
Ok, I think I get it. But, I have another question, I know that $\text{Sym}(n)$ is generated by $\{ (1, 2); (1, 3); (1, 4); ...; (1, n) \}$. But for some reason, I don't see why $\text{Sym}(n)$ is generated by $\{ (1; 2); (2; 3); (3; 4); ...; (n-1; n)\}$. Say, how can I generate $(1; 5)$ using the above generators? Thank you very much, –  user49685 Dec 11 '13 at 18:44
    
@user49685: You can change my answer to replace (2,3) with (1,3) [ both times ] and (n-1,n) with (1,n) [ both times ] to get an answer you already understand [ the rest of the answer actually doesn't change ]. To answer your question: (1,5) = (1,2)*(2,3)*(3,4)*(4,5)*(3,4)*(2,3)*(1,2); (1,n) = (1,2)*....*(n-1,n)*...*(1,2). –  Jack Schmidt Dec 11 '13 at 20:11
    
Yay, I get it, thank you vvvveeeerrryyyy much for your kind support, and explanation. :* –  user49685 Dec 12 '13 at 10:17

Yes we can.

For $A_n$, we can embed the given group in some $S_{n-2}$ and then for the additional two elements choose the identity or the transposition according as the element of $S_{n-2}$ is even or odd.

For $SL(n,q)$, we can embed the given group in some $GL(n-1,q)$ and then choose the diagonal element in the additional row and column as the reciprocal of the determinant of the element of $GL(n-1,q)$.

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I think these are backwards? Alt(3) does not embed in Sym(1), but Sym(3) embeds in Alt(5). –  Jack Schmidt Mar 15 '11 at 7:28
    
@Jack Schmidt: Sorry, I used ambiguous pronouns -- I've replaced them with the intended referents. –  joriki Mar 15 '11 at 7:31
    
Ah, much clearer now. –  Jack Schmidt Mar 15 '11 at 7:37

To your first question (embeddable within $A_n$) I think the answer is yes for obvious reasons: one can embed $S_n$ within $A_{n+2}$.

(Consider the subset of $A_{n+2}$ that stabilizes the first $n$ elements (as a set), it's obvious that this set will consist of all permutations of these elements, where it may or may not interchange the last two points, depending on the sign. For instance for $n=3$ we retrieve $S_3$ as $(1\ 2\ 3)$, $(1\ 2)(4\ 5)$, etc...)

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