# Conjugates

• November 11th 2009, 11:04 PM
davido
Conjugates
How many conjugates does the permutation (123) have in the group S<sub>3</sub> of all permutations on 3 letters? Give brief reasons for your answers.

Answers were provided for this question, but after going carefully through it, I still don't know what's going on at one point in the answer, which may prove crucial to my understanding of the problem (where I've put question marks). Below is the solution provided:

"We can use the stabilizer-orbit relationship to see that the size of the conjugacy class of (123) is equal to the index of the stabilizer. But the stabilizer is a subgroup of S3 which contains at least <(123)> [????? why]. If it were to contain more then it must be all of S3 (by Lagrange's theorem) and so contain, for example, (12). But (12)(123)(12)-1 = (132) and so the centralizer is exactly <(123)> [????]. Since it has order 3, it also has index 2 and so (123) has 2 conjugates."

Thank you.
• November 12th 2009, 01:27 AM
tonio
Quote:

Originally Posted by davido
How many conjugates does the permutation (123) have in the group S<sub>3</sub> of all permutations on 3 letters? Give brief reasons for your answers.

Answers were provided for this question, but after going carefully through it, I still don't know what's going on at one point in the answer, which may prove crucial to my understanding of the problem (where I've put question marks). Below is the solution provided:

"We can use the stabilizer-orbit relationship to see that the size of the conjugacy class of (123) is equal to the index of the stabilizer. But the stabilizer is a subgroup of S3 which contains at least <(123)> [????? why]

Because for ANY $x\in G$, for ANY group $G$, we have that $x^{-1}xx=x\Longrightarrow x$ centralizes itself, which simply means that any element in any group commutes with itself and, in fact, with any of its powers

. If it were to contain more then it must be all of S3 (by Lagrange's theorem) and so contain, for example, (12). But (12)(123)(12)-1 = (132) and so the centralizer is exactly <(123)> [????]

As seen above, (123) centralizes (123) and thus the group generated by (123), i.e. <(123>, centralizes (123)

Tonio

. Since it has order 3, it also has index 2 and so (123) has 2 conjugates."

Thank you.

.
• November 12th 2009, 01:44 AM
Swlabr
Quote:

Originally Posted by davido
How many conjugates does the permutation (123) have in the group S<sub>3</sub> of all permutations on 3 letters? Give brief reasons for your answers.

Answers were provided for this question, but after going carefully through it, I still don't know what's going on at one point in the answer, which may prove crucial to my understanding of the problem (where I've put question marks). Below is the solution provided:

"We can use the stabilizer-orbit relationship to see that the size of the conjugacy class of (123) is equal to the index of the stabilizer. But the stabilizer is a subgroup of S3 which contains at least <(123)> [????? why]. If it were to contain more then it must be all of S3 (by Lagrange's theorem) and so contain, for example, (12). But (12)(123)(12)-1 = (132) and so the centralizer is exactly <(123)> [????]. Since it has order 3, it also has index 2 and so (123) has 2 conjugates."

Thank you.

This can actually be generalised in quite a neat way:

Two permutations in $S_n$ are conjugate if and only if they have the same cycle structure.

To do this, you simply need to prove the following result, which is also very nice (and useful):

Let $i_1, i_2, \ldots i_k \in \{1, 2, \ldots, n\}$. Then for $\sigma \in S_n$ we have $\sigma^{-1} (i_1 \text{ } i_2 \ldots i_k) \sigma=(i_1 \sigma \text{ } i_2 \sigma \ldots i_k \sigma)$.
• November 12th 2009, 02:35 AM
davido
I see now, thanks so much. I didn't take much notice of "any element in any group commutes with itself and, in fact, with any of its powers", especially about the powers.