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I want to figure out how many elements of $S_5$ with the same cycle type as (12)(345).

I think the way to go about doing this would be to say :

There are $5C3=10$ ways to choose the elements in the 3-cycle and then $2C2=1$ way to choose the elements in the transposition.

As permutations are in general non-commuative there are also 2 ways to order the transposition and the 3-cycle (i.e. the transposition first or the 3 cycle first )

So in total there are $10 \times 1 \times 2=20 $ elements of this type.

Is this reasoning correct ?

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No, the order of the transposition and the $3$-cycle is irrelevant because permutations that dont move common elements commute.

The extra $2$ factor comes from the fact that $(1,2,3)$ and $(1,3,2)$ are distinct, in general $S_n$ has $(n-1)!$ $n$-cycles.

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Your 2-cycle and your 3-cycle are disjoint, so they commute.

Just because $3, 4, 5$ are in your 3-cycle does not mean your 3-cycle is $(3\,4\,5)$. It could also be $(3\,5\,4)$.

Of course, these two corrections lead to the same total number of elements of this cycle type.

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  • $\begingroup$ So are the only ways you can have say 3,4,5 in the 3-cycle , (345)=(453)=(534) and (354)=(543)=(435) ? $\endgroup$ – excalibirr May 2 '18 at 0:31
  • $\begingroup$ @exodius : There are only $3! = 6$ ways to write three symbols in an order and you've written all six of them, so yes. $\endgroup$ – Eric Towers May 2 '18 at 3:28

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