Conjugate to σ in the alternating group There is $S_7$ and the subgroup $\langle σ\rangle = \{(1 \; 2\; 3\; 4 \; 5)^i | i \in \mathbb Z\}$ of $S_7$ generated by $σ = (1 \; 2\; 3\; 4 \; 5)$

Then, how many elements are conjugate to σ in $A_7$?

The order of σ is 5 and well.. I know that conjugation of symmetric group(preserve the cycle type) but how to find the number of elements?

Thanks.
The cycle decomposition of the given permutation $\sigma=(1,2,3,4,5)(6)(7)$ contains one $5$-cycle and two fixed points ($1$-cycles).  By the splitting criterion [1,2], because the cycle lengths $5$, $1$, and $1$ are not distinct odd numbers (there are two $1$-cycles in $\sigma$), the conjugacy class of $\sigma$ does NOT split when restricted to $A_{7}$ from $S_{7}$.



The process is: (hover to view)


   There are ${7\choose 5}\cdot4!=504$ [3] conjugates of $\sigma$ in $S_{7}$.
 By the splitting criterion, because the cycle lengths of $\sigma$ (namely $5$, $1$, and $1$) are not distinct odd numbers, the conjugacy class of $\sigma$ does not split.  So there are exactly $504$ conjugates of $\sigma$ in $A_{7}$.
The cycle decomposition of the given permutation $\sigma=(1,2,3,4,5)(6)(7)$ contains one $5$-cycle and two fixed points ($1$-cycles).  By the splitting criterion [1,2], because the cycle lengths $5$, $1$, and $1$ are not distinct odd numbers (there are two $1$-cycles in $\sigma$), the conjugacy class of $\sigma$ does NOT split when restricted to $A_{7}$ from $S_{7}$.



The process is: (hover to view)


   There are ${7\choose 5}\cdot4!=504$ [3] conjugates of $\sigma$ in $S_{7}$.
 By the splitting criterion, because the cycle lengths of $\sigma$ (namely $5$, $1$, and $1$) are not distinct odd numbers, the conjugacy class of $\sigma$ does not split.  So there are exactly $504$ conjugates of $\sigma$ in $A_{7}$.