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I have heard that all finite subgroups are isomorphic to a subgroup of $S_n$. I was thinking about examples of this. In particular I would like to know how this works for certain matrix groups.

The group $M_2(\mathbb{Z}_2)$ (group under matrix addition) has order $16$. How can I make this a subgroup of $S_n$ for some $n$? Is there some concrete map? How can one do this in general?

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  • $\begingroup$ Any finite group indeed embeds in to $S_n$ for some $n$. But this is completely pointless. (in the sense that this fact is not really useful for any calculation) Ie a group of order 16 embeds into $S_{16}$. See math.uconn.edu/~kconrad/blurbs/grouptheory/gpaction.pdf maybe. $\endgroup$ – Jack Yoon May 1 '15 at 12:27
  • $\begingroup$ @JackYoon: Even though it is pointless, I would still like to see how this works for $M_2(\mathbb{Z}_2)$. $\endgroup$ – John Doe May 1 '15 at 12:28
  • $\begingroup$ The proof in the text constructs it. (theorem 1.1) $\endgroup$ – Jack Yoon May 1 '15 at 12:30
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Note that $M_2(\mathbb{Z}_2)$ is not a group under matrix multiplication. However, if we take the invertible matrices, we obtain the group $GL_2(\mathbb{Z}_2)$ of order $6$. Now it is easy to embed this group as a subgroup of $S_n$ - it is isomorphic to $S_3$.
The additive group $M_2(\mathbb{Z}_2)\simeq (\mathbb{F}_2)^4$ embeds into $S_{16}$ by Cayley's theorem.

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  • $\begingroup$ Sorry, I meant the group $M_2(\mathbb{Z}_2)$ under addition. $\endgroup$ – John Doe May 1 '15 at 12:37
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    $\begingroup$ Also imbeds to $S_8$ by sending each of the four generators to $(1,2), (3,4) (5,6)$ and $(7,8)$. (Special exception as this group is abelian so it was easy to find a smaller example). $\endgroup$ – Jack Yoon May 1 '15 at 13:27

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