I want to know if I can considere any group of order $n$ is isomorphic to one of $S_n$. Is that true? I can't find a proof.
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Yes, the group acts on itself by multiplication from the left, and this gives an injective homomorphism from the group to $S_n$. – Tobias Kildetoft Feb 26 '14 at 10:27
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5This is known as Cayley's theorem: http://en.wikipedia.org/wiki/Cayley%27s_theorem – dani_s Feb 26 '14 at 10:29
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Thank you. Now is clear. I didn't know this theorem. – MaríaCC Feb 26 '14 at 10:38
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(To remove this from unanswered questions)
Yes, this is known as Cayley's theorem.
Label the elements of $G$ as $g_1$, $g_2$, up to $g_n$. For each $g_i$ define a permutation $\pi_i$ such that if $g_i \cdot g_j = g_k$ then $\pi_i(j) = k$. Since multiplication is invertible in a group, $\pi_i$ is a bijection, and since multiplication is closed, $\pi_i$ is a permutation. If $\pi_i( \pi_j(k)) =\pi_i(l) = m$, then $g_j \cdot g_k = g_l$ and $g_i \cdot g_l = g_m$ so $(g_i \cdot g_j) \cdot g_k = g_m$. Hence if $g_i \cdot g_j = g_p$, then $\pi_i \circ \pi_j = \pi_p$. Thus the function $g_i \mapsto \pi_i$ is a homomorphism. It is clearly injective, since if $g_1$ is the identity of $G$, then $\pi_i(1) = i$ identifies each $\pi_i$; that is, the inverse is $\pi \mapsto g_{\pi(1)}$. Hence we have an isomorphism of $G$ with a subgroup of $S_n$.
The subgroups of this form are called regular subgroups.
One can also include the automorphism group by taking a normalizer (this gives the holomorph). Recognizing such holomorphs can be very useful, and often described as permutation groups with a regular normal subgroup, such as the affine groups.
Jack Schmidt
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