I am given the permutation $(13)(24)$ in the symmetric group $S_4$.
I've been asked to find all the conjugates in this symmetric group, though I am confused as to what the conjugate would be for a permutation. I understand that the inverse is simply the reverse of the direction of the permutation (which would be the same here, I believe), but for the conjugate I am at a loss. I know for example that the conjugate of a real number is the same, and for a complex number it is a sign change. Would appreciate some help.
Given elements $a$ and $b$ in a group $G$, $a$ is said to be conjugate to $b$, if there is some $c \in G$ satisfying $c^{-1}ac = b$.
Now, conjugacy is an equivalence relation on elements of a group, because :
$a = eae=e^{-1}ae$, where $e$ is the identity.
If $a = c^{-1}bc$, then $b = (c^{-1})^{-1} a c^{-1}$.
If $a = d^{-1}bd$ and $b = f^{-1}cf$ then $a = (fd)^{-1}c(fd)$.
Hence, the equivalence class of $a$ i.e. the set of elements which are conjugate to $a$, are called the conjugates of $a$.
For $S_n$, we have the cycle notation :for example $(abc)$ is the permutation taking $a \to b, b\to c, c\to a$. Every element can be written as a product of disjoint cycles. Now, let $h$ be a permutation in $S_n$, and $g = (a_1...a_i)(b_1...b_j)...(t_1...t_l)$ be some element of $S_n$ written as a product of disjoint cycles.
Then, turns out $h^{-1}gh $ has a cycle notation : $(h(a_1) ... h(a_i))(h(b_1)...h(b_j))...(h(t_1)...h(t_l))$ as a product of disjoint cycles. Therefore, $h^{-1}gh$ retains the disjoint cycle structure of $g$ : the same number of cycles, and each cycle has the same number of elements.
Conversely, given any permutation $f$ with the same disjoint cycle structure as $g$, then a $h$ will exist so that $h^{-1}gh = f$. Write down $f$ like how I wrote down $g$ and see how $h$ can be described.
From the above paragraphs follows :
Hence, all you need to do, is to list permutations, which have the same cycle structure as $(13)(24)$, which means that it is of the form $(ab)(cd)$, where $\{ a,b,c,d\} = \{ 1,2,3,4\}$.
So, in this case, you would get permutations that look like this:
$(12)(34)$
$(13)(24)$
$(14)(23)$
The above three are conjugate, and no other permutation is conjugate to any of these.