1. ## Conjugacy Classes

Just a little confused about conjugacy class equations.

Example:

In the symmetric group $S_5$ how many elements are there in the conjugacy class of the element $(12)(34)$?

I get that this is the amount of different possible ways to have a possible combination of 2 2 1 in $S_5$, and I know the answer's 15, I'm just not sure how to calculate this.

2. I've found how this is calculated, but just not sure where they actually get these formulas from:

Symmetric group:S5 - Groupprops

3. Solved it. If anyone's reading this later on the following video explains it nicely!

4. Originally Posted by craig
I've found how this is calculated, but just not sure where they actually get these formulas from:

Symmetric group:S5 - Groupprops
The link you gave doesn't seem to work at the moment...

Anyway, for permutations of the form (ab) you have 5 choices for a and 4 for b (as (aa) is not a permutation). However, (ab)=(ba) so you have to account for this, so you have 5.4/2=10 choices.

Similarly, for permutations of the form (123) you have 5.4.3/3=20 choices, as (abc)=(bca)=(cab), and so it goes on for the other basic cycles.

Now, for (abc)(de) things start to get a bit more interesting. You have 5.4.3.2.1/3.2=20 choices, calculated the same as above.

Adding all these up (and remembering your identity element) you should get 10+20+30+24+20+1=105. That means you have 15 till to find.

Therefore, the conjugacy class of permutations of the form (ab)(cd) has 15 elements. This is proven properly in the following way:

(ab)(cd) has 5.4.3.2 basic choices, as mentioned earlier. However, you need to divide through by 2, because (ab)(cd)=(ba)(cd), and again by two because (ab)(cd)=(ab)(dc). Finally, notice that (ab)(cd)=(cd)(ab). So you need to divide through by 2 again.

Thus, you have 5.4.3.2/2.2.2=5.3=15, as required.

Does that make sense?

5. Thanks for the reply! Yeh that makes sense, it was the final dividing through by 2 that threw me!

Oh and the link seems to work fine for me?

Cheers again for help!

6. Originally Posted by craig
Thanks for the reply! Yeh that makes sense, it was the final dividing through by 2 that threw me!

Oh and the link seems to work fine for me?

Cheers again for help!
Yeah-it works fine for me too now! Don't know what was up - it took me to some page apologising for something and promising to right it asap. I didn't read it in detail though...

7. the way i've always counted them is like this:

by insisting that i < j, we can list the traspositions in an orderly fashion, like so:

(1 2), (1 3), (1 4), (1 5), (2 3), (2 4), (2 5), (3 4), (3 5), (4 5), making it clear there are 4+3+2+1 = 10 transpositions.

now suppose we want to count how many (a b)(c d) there are, with (a b)(c d) disjoint.

clearly, we have 10 "free" choices for (a b). the next choice must be a transposition that has neither a nor b in it, there are 4 transpositions with "a" in it, and 4 transpostions with "b" in it. but of the transpositions with "a" in it, one of those is (a b), so we must exclude only 4+3 = 7 of the 10 transpositions to choose from. that gives us exactly 3 transpositions left that have neither a nor b in it. for example, if a = 1, b = 2, the 3 transpositions are (3 4), (3 5) and (4 5).

that gives 30 possible ways to create a disjoint pair of 2-cycles (a b)(c d). since we may have also chosen (c d)(a b) this way, we must divide by 2, leaving 15 possible disjoint pairs of 2-cycles.

8. Originally Posted by craig
Just a little confused about conjugacy class equations.

Example:

In the symmetric group $S_5$ how many elements are there in the conjugacy class of the element $(12)(34)$?

I get that this is the amount of different possible ways to have a possible combination of 2 2 1 in $S_5$, and I know the answer's 15, I'm just not sure how to calculate this.

Barring the video you mentioned it doesn't seem like anyone has mentioned this. On my blog here I prove that then number of elements in the conjugacy class associated to the cycle type $1^{m_1}\cdots n^{m_n}$ in $S_n$ is $\displaystyle n!\prod_{j=1}^{n}\frac{1}{(m_j)!j^{m_j}}$ so that in your case when you've got the cycle type $1^{1}2^{2}3^04^05^0$ the number of elements of that conjugacy class is $\displaystyle 5!\cdot\frac{1}{1!1^1 (2!)2^2}=\frac{120}{8}=15$