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How do I figure out how many elements are in each conjugacy class of the symmetric group S10, for the class (****) star being any element, i get we should fix one number so i decided to choose 1. i then realised there are 3!ways of choosing the other 4 elements but not sure what to do next.

user310686
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  • Possible duplicate of http://math.stackexchange.com/questions/449041/counting-the-number-of-elements-in-a-conjugacy-class-of-s-n?rq=1 – Captain Lama Apr 19 '16 at 19:45

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Well, no, there are 10 choices for the first number, then 9 for the second, 8 for the third, and 7 for the fourth.

This appears to make for $$10 \cdot 9 \cdot 8 \cdot 7$$ $4$-cycles, but you have overcounted, because a 4-cycle can be written in 4 different ways, e.g. $$ (1234) = (2341) = (3412) = (4123). $$ So you get $$ \dfrac{10 \cdot 9 \cdot 8 \cdot 7}{4} $$ 4-cycles.

Andreas Caranti
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  • okay thankyou, so is similar method used for a 2 order cycle next to 5 – user310686 Apr 19 '16 at 20:22
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    Yes. It gets more complicated when you have two (or more) cycles of the same length, because then you have to take into account the possibility of swapping them. For instance, if you are counting the permutations that are the product of $5$ disjoint $2$-cycles, you will obtain the number $\dfrac{10!}{(2!)^{5} \cdot 5!}$. – Andreas Caranti Apr 20 '16 at 10:57