Let G=Sym(n) .Let C be a cyclic group. Suppose that $\displaystyle \theta \in Hom(C, G). $then how to show that the generators for $\displaystyle \theta (C)$ in G all have the same cycle type and so are conjugate in G.

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- Mar 20th 2009, 12:43 AMdimukCyclic groups
Let G=Sym(n) .Let C be a cyclic group. Suppose that $\displaystyle \theta \in Hom(C, G). $then how to show that the generators for $\displaystyle \theta (C)$ in G all have the same cycle type and so are conjugate in G.

- Mar 20th 2009, 08:36 AMThePerfectHacker
I am assuming that the cyclic group is finite.

If $\displaystyle a$ generates $\displaystyle \theta (C)$ then $\displaystyle a^j$ generates $\displaystyle \theta (C)$ where $\displaystyle j$ is relatively prime with the order of $\displaystyle a$ because a homomorphic image of a cyclic group is cyclic. Let $\displaystyle a=\sigma_1\sigma_2 ... \sigma_k$ for disjoint cycles $\displaystyle \sigma_i \in S_n$. The order of $\displaystyle a$ is the least common multiple of the lengths of all these cycles. Therefore, $\displaystyle |\sigma_i|$ divides $\displaystyle |a|$ for all $\displaystyle i$. Now if $\displaystyle j$ is an integer so that $\displaystyle (j,|a|) = 1$ then certainly $\displaystyle (j,|\sigma_i|)=1$. Therefore, if $\displaystyle \sigma_i$ is an $\displaystyle i$-cycle then $\displaystyle \sigma_i^j$ still stays as an $\displaystyle i$-cycle. Thus, we have $\displaystyle a^j = \sigma_1^j ... \sigma_k^j$ where $\displaystyle \sigma_i$ is same cycle length as $\displaystyle \sigma_i^j$ which means $\displaystyle a$ and $\displaystyle a^j$ have the same cycle type and consequently they are conjugate.