# Verifying the calculation of the number of conjugate elements

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?

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.
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 $$\mathrm{S_7)$$