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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 $n,q\in \mathbb{N}$?

  • 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 Answers3

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Yes.

The symmetric group $Sym(n)$ is generated by $\{(1,2), (2,3),\ldots, (n−1,n)\}$. 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 $\pi\in Sym(n)$ and sends it to $\pi⋅(n+1,n+2)^{\text{sgn}(\pi)}$, where $\text{sgn}(\pi)\in\{0,1\}$ is the parity of the permutation.

In other words, $G\le Sym(n)\le Alt(n+2)$ 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, $G\le GL(n,q)\le SL(n+1,q)$ embeds any group into a (slightly larger) special linear group.

Peter Phipps
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Jack Schmidt
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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
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    @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
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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 $GL(n-1,q)$.

joriki
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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...)

Myself
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