Verifying the calculation of the number of conjugate elements

Question

I would like to verify my understanding:

Consider action of $S_7$ on itself by conjugation. I'm trying to compute:

1. $Stab_{S_7}((1 2))$
2. $Stab_{S_7}((1 2 3 4 5 6 7))$
3. $Stab_{S_7}((1 2 3)(4 5 6))$

I'm following this rule: two elements are conjugate if and only if they have the same cycle type.

For the first one it should be $(x_1x_2)$ so I choose 2 elements from 7 and get the orbit: $\binom{7}{2}$ meaning $Stab_{S_7}((1 2))=\frac{7!}{\binom{7}{2}}$

For the second one it should be cycle type $(7)$ so there is only one cohse meaning $Stab_{S_7}((1 2 3 4 5 6 7))=7!$

For the third one it should be cycle type $(3,3)$ so we choose 3 elements from 7 for the first and organize in circle with $\binom{7}{3}\cdot 2!$ possibilities. Then we choose $3$ from $4$ for the second one and organize in circle with $\binom{4}{3}\cdot 2!$. All that we should divide by $2$, so we get:

$$\frac{\binom{7}{3}\binom{4}{3}\cdot 4}{2}=\binom{7}{3}\binom{4}{3}\cdot 2$$

And then: $Stab_{S_7}((1 2 3)(4 5 6)) = \binom{7}{3}\binom{4}{3}\cdot 2$.

I feel like it isn't correct. If so, how to solve it?

Answer 1

The first one is correct. For the second one, there are many cycles of $7$ elements. Thinking of it as a permutation $f$, you have $6$ choices for $f(1)$, $5$ choices for $f(f(1))$, $4$ choices for $f(f(f(1)))$, and so on until the last thing must map to $1$, so there are $6!$ cycles of length $7$, hence $\mathrm{Stab}_{S_7}((1234567))=7!/6!=7$.

For the third one, you are correct in counting the number of permutations of cycle type $(3,3)$, but that is not the same as $\mathrm{Stab}_{S_7}((123)(456))$. As you do in the other ones, you have to take that value and divide $7!$ by it.

Answer 2

I will attempt to do the 3rd one for you: According to the type formula, we know that the conjugate of (123)(456) is some sort of double 3 cycle. Note that the stabilizer here is heavily determined: If i send 1 to 5, for example, then i must send 2 to 6 and 3 to 4. I then have a choice of whether i send 4 to 1,2 or 3. By using this property, we can reason as follows: 1. I can send 1 to 6 possible things(excluding 7). Each choice also determines where i send 2 and 3. 2. After having sent 1, i now have 3 choices on where to send 3(once again cant send to 7). So we have in total 18 choices, and each one corresponds to a different permutation of $$\mathrm{S_7)$$