In my lectures on Group Theory our lecturer claimed the following: All 3-cycles are pairwise conjugate.
He then went on to prove this but I am struggling with understanding his proof. I will try and lay out what I understand here and hopefully somebody can help me to fill in the gaps.
Proof:
Note that $A_{n} = \frac{n!}{2}$.
Consider $\sigma = (1 \ 2 \ 3) \in A_{n}$ and consider $N(\sigma) := \{\tau \ | \ \tau \sigma \tau^{-1} = \sigma, \ \tau \in A_{n} \}$. Then we can say that $\forall \tau \in N(\sigma)$ it holds that $(\tau(1) \ \tau(2) \ \tau(3)) = \sigma$.
Therefore we can write an arbitrary $\tau$ in the following way:
$\tau = (1 \ 2 \ 3)^{\lambda} \circ \ $(some even permutation of $\{4,5,...,n\}$).
Hence, as there are three choices for $(1 \ 2 \ 3)^{\lambda}$, and $\frac{(n-3)!}{2}$ choices for some even permutation of $\{4,5,...,n\}$, we can see that #$N(\sigma)=3\frac{(n-3)!}{2}$.
Using the Orbit-Stabiliser Theorem it holds that #$A_{n} =$ #$N(\sigma) \cdot$#$C(\sigma)$. [Note: I think this theorem wasn't very well explained in our lectures, I believe $N(\sigma)$ is the stabiliser of $\sigma$ and $C(\sigma)$ is the orbit of $\sigma$]
Therefore #$C(\sigma) = \frac{\frac{n!}{2}}{3\frac{(n-3)!}{2}} = \frac{1}{3} \cdot \frac{n!}{(n-3)!} = 2\frac{n!}{3!\cdot(n-3)!} = 2 \cdot \binom{n}{3}$.
[Note: this is where I get very confused and have no real idea of the conclusions the professor was able to draw from this]
Now, in general, how many elements are there in the set of 3-cycles in $A_{n}$?
#{3-cycles in $A_{n}$} = $2 \cdot \binom{n}{3} = $#$C(\sigma)$
Therefore {3-cycles in $A_{n}$} $\subseteq C(\sigma)$ and as the number of elements in each set is the same, these two sets are equal.
[Note: This is the end of the proof and I really don't see how this proves that all 3-cycles are pairwise conjugate. Any help would be greatly appreciated]