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Question Prove that any group G of order 33 is cyclic, by considering the conjugation action of a subgroup of G on G.

This is a repeat question, I really do not understand how I am supposed to use conjugation action the show this. In a previous part of the question i was asked to prove the orbit stabiliser theorem, so that might be relevant, but I’m not sure.

My thought process so far: due to Cauchys theorem G has an element of order 3, and an element of order 11. I believe there’s only 1 subgroup of order 11, which would be isomorphic to $C_{11}$, so all 10 non-identity elements there have order 11. I was trying to consider the conjugation action of this on G, but I wasn’t having much luck.

Anon314159
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  • Without Sylow, $G$ has an element of order $3$ and one of order $11$. This is by Cauchy's theorem, which you should have covered already. The subgroups $N$ and $H$ generated by these elements are normal subgroups isomorphic to $C_3$ and $C_{11}$. Since $3$ and $11$ are coprime,. it is easy to see that $G\cong C_3\times C_{11}$ is cyclic. – Dietrich Burde May 17 '24 at 13:13
  • The duplicates also inlcude answers without Sylow, by the way. – Dietrich Burde May 17 '24 at 13:15

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