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If I have a symmetric group on $9$ letters, how many elements in the group have order $7$?

I’ve started off thinking about the disjoint cycles, so in cyclic form the only elements can be $7,1,1$ to get order $7$ (as sum of cycles has to equal $9$ but lcm has to equal $7$).

So I started to think about permutations of it. And I’m stuck there.

egreg
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L G
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2 Answers2

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You're definitely on the right track. Now you have to count the number of different $7$-cycles in $S_9$. You can do that by first counting how many $7$-element subsets there are in $S_9$, and for each of those subsets, how many different cycles you can make from those elements.

Arthur
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  • I’ve looked at permutations of 7 so 7! Multiplied by 3! Due to the different orders you can rearrange the 7,1,1 to get 30240 but that definitely seems wrong – L G Oct 29 '18 at 15:14
  • @LG That's not quite right, no. For the permutations of $7$, yes, there are $7!$ of them, but some of them represent the same cycle. So you need to take care of that somehow (for each cycle, how many different permutations represent that cycle?). As for the whole $7, 1, 1$ deal, that's not what $7, 1, 1$ means. It means you need to pick $7$ elements out of the $9$ to put in the $7$-cycle, and the remaining two elements will be in separate $1$-cyles. – Arthur Oct 29 '18 at 15:25
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You want to count how many different seven element sequences you can get from nine: $9$ choices for the first element, eight for the second and so on.

However two seven element sequences may represent the same cycle. How many sequences represent a given cycle?

egreg
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