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I've bolded the incorrect part.

Find the number of order $3$ elements in $S_6$

My "Solution":

There are two ways to obtain these, either a 3-cycle or product of 2 disjoint 3-cycles.

Counting 3-cycles:

Choose 3 of 6 elements, this will determine the first 3 elements. We don't worry about the order of the rest. To get the distinct cycles (noting $(1 2 3) = (2 3 1)$), we fix the first element and permute the remaining elements. Thus: $\frac{6!}{3!3!} (3-1)!=40$

Counting Disjoint 3-cycles:

Choose 3 of 6 elements, this will determine the first 3 elements. Using the above computation we obtain $40$. We need to account for of the rest. Noting again that some cycles are similar, we fix the first element and permute the remaining to get distinct cycles, thus: $40\times (3-1)! = 80$

Thus: $120$ elements of order $3$.

The bolded reasoning is incorrect. The answer key says it should be 40. How am I over counting?

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  • $\begingroup$ You have two 3-cycles. You could the permutation twice starting with each of these. $\endgroup$ Mar 20, 2018 at 18:36
  • $\begingroup$ Use $\times$ for $\times$. $\endgroup$
    – Shaun
    Mar 20, 2018 at 18:37
  • $\begingroup$ @LordSharktheUnknown Could you elaborate? I don't quite follow $\endgroup$
    – yoshi
    Mar 20, 2018 at 18:39

3 Answers 3

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I like to think in terms of automorphism groups. That is, think of how many permutations of the symbols $1,\dots,6$ fix the permutation

$$ (123) = (123)(4)(5)(6). $$

So the question is: if we create a 3-cycle by taking a permutation of $6$ such as $312546$ and grouping together the first three elements to make $(312)$, how many times to we get the same 3-cycle?

With $(123)(4)(5)(6)$, we see that there are $3!$ ways to permute $4,5,6$ and they don't affect the 3-cycle we get. There are a further $3$ ways to permute $(123) = (231) = (312)$ without changing the 3-cycle. Thus there are

$$ \frac{6!}{3!3} = 40$$

3-cycles.

For products of two 3-cycles, we ask the same question but for the permutation

$$ (123)(456). $$

Again there are $3$ ways to permute $1,2,3$ and $3$ ways to permute $4,5,6$ without changing the permutation. But we can also exchange $1,2,3$ with $4,5,6$ to create $(456)(123)$. This gives us another factor of $2$ which you missed.

In total there are

$$ \frac{6!}{3\cdot 3 \cdot 2} = 40 $$

products of two disjoint 3-cycles.

With your solution, just note that you are counting $(123)(456)$ and $(456)(123)$ twice and divide by 2 to fix.

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Two elements of $S_{n}$ are conjugate if and only if they have the same cycle structure. There are two possible cycle structures for an element of order $3$ in $S_{6}$, as you noted yourself. Hence we need to count the conjugates of $x = (123)$ and of $y = (123)(456)$ in $S_{6}$ ( and add these numbers to get the number of elements of order $3$ in $S_{6}$). Now $C_{S_{6}}(x)$ has order $18$ (it's isomorphic to $\langle x \rangle \times S_{3}$) so $x$ has $6!/18 = 40$ conjugates, and $C_{S_{6}}(y)$ also has order $18$ (it's isomorphic to $C_{3} \wr C_{2},$ which is in fact isomorphic to $C_{3} \times S_{3}$), so $y$ also has $40$ conjugates. The key point to note is that the element $(14)(25)(36)$ commutes with $(123)(456) = y,$ which gives a slightly less obvious element of $C_{S_{6}}(y)$ and in one sense explains why the reasoning in bold in the question statement is incorrect (since there you get twice as many elements there as you should).

Hence $S_{6}$ contains $80$ elements of order $3$ in total, as already noted above by a different method.

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We only have to consider the following Young-Tableaus: $$ \begin{align*} [a_1][a_2][a_3]\\ [a_4]\\ [a_5]\\ [a_6] \end{align*} $$ and

\begin{align*} [a_1][a_2][a_3]\\ [a_4] [a_5] [a_6] \end{align*}

The first Young-Tableau can be filled in $6!$ possible ways by the $6$ objects. However, cyclic shifts in the first row don't matter and permuting the last three rows also doesn't matter, so there are $6!/(3*3!) = 40$ ways.

The second Young-Tableau can be filled in $6!$ possible ways as well, but then there are $3$ cyclic shifts in each row as well as $2$ ways to order the rows. So we get $6!/(3*3*2) = 40$ as well.

In total, this gives $40+40=80$ elements of order 3.

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