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I want to use Cayley's theorem to determine a subgroup in $S_n$ ( for n as small as possible) which is isomorphic to $C_{5}$.

I believe this subgroup to be $ \langle (1 2 3 4 5) \rangle $. Here is my reasoning:

$C_5 = \{ e, g, g^2, g^3, g^4\}$ is generated and hence determined by $g$. The isomorphism $ \rho$ used in the proof of Cayley's theorem maps $g$ to $(1 2 3 4 5)$ in $S_5$. But since $ \rho $ is a homomorphism, $ \rho(C_5) = \langle (1 2 3 4 5) \rangle.$ Therefore, $C_5 \cong \langle (1 2 3 4 5) \rangle.$

Is this correct? Any help is greatly appreciated!

  • For the "$n$ as small as possible" part see https://math.stackexchange.com/questions/191446/efficient-version-of-cayleys-theorem-in-group-theory – Arnaud D. May 21 '19 at 12:03
  • It's clear that $S_5$ is minimal for this. If $n<5$ then $5,\nmid ,n!$ – lulu May 21 '19 at 12:13

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I don't see what Cayley's theorem has to do with anything; it gives you an embedding of your group into some symmetric group, but gives you no guarantees of that being the smallest symmetric group.

In stead, a quick check will show that $S_n$ has no elements of order $5$ for $n<5$. And you have found a subgroup of order $5$ in $S_5$, hence $n=5$ is the smallest $n$ for which such a subgroup exists. Done.

Servaes
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  • Thanks! I see how this quickly gives the least $n$. But, regardless, is my reasoning correct? –  May 21 '19 at 11:48
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    @wittbluenote Your reasoning shows that $S_5$ has such a subgroup. You have yet to show that $n$ is as small as possible. – Servaes May 21 '19 at 11:50
  • I understand now, thank you! –  May 21 '19 at 11:51