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The elements in $$S_{n}$$ have the following orders: 1,2,3,...,n I'm thinking about a specific case of $$S_{7}$$ Now, to find how many elements of order 6, for example, I look at the following permutations: $$\alpha = (a b c d e f),$$ $$\beta = (a b)(c d e),$$ $$\gamma = (a b)(c d e)(f)$$

For $\alpha$, there are 5! elements, for $\beta$, there are $\frac{5*6}{2}*\frac{3*4*2}{3} = 120 $ etc. My question is, do I have to count $\gamma$ ? Because if I proceed in the same way, I get more elements than 7!

JMK
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1 Answers1

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When you write an element of $S_7$ like $$ (abcdef) $$ in cycle notation, you're using fewer than $7$ symbols, so there's one element which you aren't explicitly saying how you're permuting. (For example, if your cycle is $(124567)$, you aren't saying what happens to $3$.)

Part of the convention of cycle notation that each symbol you don't write is mapped to itself: that is, $$ (abcdef)=(abcdef)(g) $$

So you don't need to worry about cycle types with $1$-cycles in them: they've already been taken care of. That is, your $\beta$- and $\gamma$-type cycles are identical to each other. (Or, equivalently, you could worry about cycle types with $1$-cycles in them, but then you should always be using seven symbols to describe them; e.g., you could ask about cycle types $$ \alpha=(abcdef)(g)\\ \beta=(abc)(de)(f)(g) $$ and so on.)

Micah
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