Find all $\tau$ in $S_4$ such that $\tau (12)(34) \tau^{-1} = (12)(34)$ I know there should be 8 but I have only found these 6 so far, so I am missing 2:
$(1)$
$(1 2)$
$(3 4)$
$(12)(34)$
$(13)(24)$
$(14)(23)$
I know that if $\sigma$ is a $k$-cycle  and $\tau\sigma\tau^{-1} = \sigma$ then $\tau$ is a power of $\sigma$. But what can we say about $\tau$ when sigma is a product of disjoint cycles?
 A: In extended notation, the permutations $\tau\in S^4$ such that 
$$\tau(1,2)(3,4)\tau^{-1} = (\tau(1),\tau(2))(\tau(3),\tau(4))=(1,2)(3,4)$$ clearly are
$$ \left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 1 & 2 & 4 & 3\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 3 & 4\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3\end{smallmatrix}\right)$$
$$ \left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 4 & 3 & 1 & 2\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1\end{smallmatrix}\right)$$
one just have to decide if the sets $\{1,2\}$,$\{3,4\}$ get switched or not, then if the action of the permutation on $\{1,2\}$ and $\{3,4\}$ is order-preserving or not. Three independent binary choices, hence eight chances.
The eight elements above form a non-abelian subgroup of $S_4$ with $8$ elements and exactly two elements of order $4$, thus a group isomorphic to the dihedral group $D_8$.
A: Given a cycle $\sigma = (a_1,\dots,a_r)$ the permutation $\tau \sigma \tau^{-1}$ is the cycle $(\tau(a_1), \dots, \tau(a_r))$. In particular it is not true that $\tau \sigma \tau^{-1} = \sigma$ only when $\tau$ is a power of $\sigma$.
If $\sigma$ is a product of cycles such as $(12)(34)$ then we can do the following:
$$ \tau (12)(34) \tau^{-1} = \tau (12) \tau^{-1} \tau (34) \tau^{-1} = (\tau(1), \tau(2)) (\tau(3),\tau(4)). $$
We can achieve this by doing any subset of the following:


*

*switching $1$ and $2$

*switching $3$ and $4$

*switching $1$ and $3$ and $2$ and $4$


The just looking at subsets of the first two bullets gives you the first 4 permutations in your list.
The third bullet by itself gives you the fifth.
Your last permutation is all three bullets together:
$$ (13)(24)(12)(34) = (14)(23) $$
The missing two permutations are given by combining the first and third and the second and third bullets:
$$ (13)(24)(12) = (1423) $$
$$ (13)(24)(34) = (1324) $$
