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Show that every finite group of order n is isomorphic to a group consisting of n x n permutation matrices under matrix multiplication.

(A permutation matrix is one that can be obtained from an identity matrix by reordering its rows)

I can't even understand the question. I think a group of n x n permutation matrices is not isomorphic to finite group of order n. I think it is isomorphic to Sn.

Help me plz!

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    $\begingroup$ Do you know Cayley's theorem? $\endgroup$ – Ayman Hourieh Jul 13 '13 at 22:11
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    $\begingroup$ If you can realise $S_n$ as a group of permutation matrices, then you can realise any subgroup of $S_n$ as such a group $\endgroup$ – Mark Bennet Jul 13 '13 at 22:12
  • $\begingroup$ The question might be rephrased as showing every finite group of order $n$ is isomorphic to a subgroup of all $n \times n$ permutation matrices. As the OP suspects, this larger group is "isomorphic to $S_n$" as every permutation of $n$ items is represented. $\endgroup$ – hardmath Jul 13 '13 at 22:14
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If you can realise $S_n$ as a group of permutation matrices, then you can realise any subgroup of $S_n$ as a subgroup, which happens to be a group of permutation matrices.

You get a representation of $G$ as a subgroup of $S_n$ by associating each element $g\in G$ with the permutation of elements $h\in G$ which sends $h\to hg$

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  • $\begingroup$ This is precisely Cayley's theorem. $\endgroup$ – user174708 Mar 13 '15 at 13:19
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The group of all permutation matrices may be $S_n$, but the problem statement asks about a group of (some) permutaion matrices. You get this immediately by embedding $G\to S_n$, and get that by considereing the action of $G$ on its underlying set by left multiplication.

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