Question:
Let $n \geq 5$ an integer, prove that the $3$-cycles of $A_n$ are pairwise conjuguates in $A_n$. Is this true for $n=3$ or $n=4$? Explain your answer.
I have a doubt concerning my conclusion in "-3" is it correct? Mostly because I don't see where we use the fact that $n \geq 5$ in my answer (so it seems to be correct for all $n$) But this is not the case.
So jump directly to my part "3-" it if you do not have time.
My Answer:
1-
First of all let remind the definition of conjuguate permuation: Two permutations $\sigma, \sigma' \in S_n$ are conjuguate if $\exists \tau \in S_n$ s.t. $ \sigma ' = \tau \sigma \tau^{-1} $. Conjugation is a way to transform one permutation into another using a third permutation and its inverse.
Exemple: With $S_4$, $\sigma = (123), \sigma'=(124)$. We are looking for $\tau \in S_4$ s.t. $(234)= \tau (123) \tau^{-1}$ and $\tau = (1243)$
So here we are asked to prove that $ \forall \sigma, \sigma' \in A_n \Rightarrow \exists \tau \in A_n$ s.t. $ \sigma' = \tau \circ \sigma \circ \tau^{-1}$.
2-
Now let's chose $\sigma = (a;b;c), \sigma'=(d;e;f) ; a,b,c,d,e,f \in \left \{1;2;...;n \right \}$ any 3-cycles in $A_n$.
Because we know that in $S_n$ it is verify $\sigma \circ ( a; b ) \circ \sigma^{-1}= (\sigma(a); \sigma (b) )$. On an other side we have $(a;b;c)=(a;b)(b;c)$.
Hence we can writte $ \tau \circ (a;b;c) \circ \tau^{-1} = \tau \circ (a;b) \circ (b;c) \circ \tau^{-1} = \tau \circ (a;b) \circ \tau^{-1} \circ \tau \circ (b;c) \circ \tau^{-1} = (\tau(a); \tau(b)) \circ (\tau(b); \tau(c))= (\tau(a); \tau(b); \tau(c) ) $.
3-
Thus it is easy to define $ \tau(a) = d , \tau(b)=e, \tau(c)=f $, hence we have shown that for every $ \sigma, \sigma'$ of 3 cycles we can define a $\tau$ such that $ \sigma ' = \tau \sigma \tau^{-1}$. And this $\tau$ is of the form thus it is in $A_n$
Q.E.D.
4-
- For $n=3 \Rightarrow A_3= \left \{ (123);(132) \right \} $ thus it is obvious that there isn't even enough elements in $A_3$ in order to be able to writte $ \sigma' = \tau \sigma \tau^{-1} $ with $ \sigma , \tau , \sigma' \in A_3 $ (all of them differentes permutations).
- For $n=4$ we have that the only permutation that sent $(123)$ to $(2;1;3)$ is $(1;2) \notin A_4 $
I have a doubt concerning my conclusion in "-3" is it correct? Mostly because I don't see where we use the fact that $n \geq 5$ in my answer (so it seems to be correct for all $n$) But this is not the case.
Thank for your help.
Below an explanation on why my prove is only correct for $n \geq 5$ .
i-
In my prove above in part "-3" I say that a permutation $\tau$ that solves this question has to verify $(\tau(a)=d; \tau(b)=e; \tau(c)=f )$ and the other elements of $S_n$ are unchanged but basically in $S_n$ there is $(n-3)!$ permutation that satisfy this condition, but by the question the permutation HAS to be too in $A_n$ that means that the correct permutation has to have an even number of permutation.
ii-
If $\tau \in A_n $ all ready so we have finished.
But if $\tau \notin A_n$ this means that $\tau$ contains an odd number of permutation. We can build an other valid permutation $\tau'$ this way $\tau' = (i;j) \circ \tau $ with $i,j \in S_n - \left \{ d;e;f \right \} $. Hence has we ve had an other permutation to $\tau$ the number of permuation is now even $ \Rightarrow \tau' \in A_n$. Moreover as $\tau'$ does not affect the 3-cycle permutation this does not modify the rest of our prove.
iii-
Now note that in the case of $n=3,4$ there is a unique permutation in $S_n$ satisfying the condition in "3-" in my question above (indeed $(4-3)!=1=(3-3)!$).
And obviously in order for $ \sigma $ to be in $A_n$ in the case of $n=3$ or $n=4$ we have that this unique permutation has to be a 3-cycle because in the other case we can not "correct" this unique permutation as we ve explained in "ii-" by the pigeon hole principle .
Exemple: As we explained only the permutatio $(12)$ verifies $(213)=(123)(12)(123)^{-1}$. Now obviouslt $(12) \notin A_n$ but we cannot correct $(12)$ by linking it with an other permutation $(i;j)$ because those $i,j$ should have to verify $i,j \neq 1,2,3 $ but still $i,j \in \left \{ 1,2,3,4 \right \} $ and this is not possible.