0
$\begingroup$

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]

$\endgroup$
  • 1
    $\begingroup$ This can't be true as in the given "proof": $(123)$ and $(132)$ are not conjugate in $A_4$. See math.uconn.edu/~kconrad/blurbs/grouptheory/conjclass.pdf. But all 3-cycles are conjugate in $S_n$ because two permutations are conjugate iff they have the same cycle structure. $\endgroup$ – lhf Jul 21 '16 at 14:05
  • $\begingroup$ Ah yes, that's very true. Thank you, maybe I misunderstood my lecture notes. $\endgroup$ – Daniel Ward Jul 21 '16 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.