Consider you have a even permutation $p$ ,which is represented by disjoint cycles and I need to multiply by three-cycle to reach identity permutation.
I have been able to solve this question when the permutation only has three-cycles or two-cycles ,that is if it has a three-cycles $(ijk)$ then I would multiply it by $(kji)$ (that is $(ijk)(kji)$ is idenity)
and if has a two-cycle $(ij)$ then find other two-cycle $(kl)$ then
form a identity $(ij)(kl)=(ijk)(jkl)$ and then apply the rule from three-cycle, but if it has a cycle of length $k$, where $k\gt 3$ I do not know the method
How should I proceed for a general case?
Best Answer
Of course we need $n > 2$.
Assume $n\ge 3$.
Claim:$\;$Every even permutation in $S_n$ can be multiplied by a product of $3$-cycles to reach the identity.
Proceed by induction on $n$
For $n=3$, the claim can be verified by direct computation.
Let $n\ge 4$ and suppose the claim holds for the case $n-1$.
Let $F_{n-1}$ be the subgroup of $S_n$ which fixes $n$ (regarding elements of $S_n$ as permutatons of the set $\{1,...,n\}$).
We can identify $F_{n-1}$ with $S_{n-1}$, hence we can freely regard elements of $F_{n-1}$ as elements of $S_{n-1}$, and vice-versa.
Let $g\in S_n$ be an even permutation.
Let $h=gc$, where $c$ is a $3$-cycle such that $c(n)=g^{-1}(n)$.
Then $h(n)=g(c(n))=g(g^{-1}(n))=n$, hence $h\in F_{n-1}$.
Since $c$ is a $3$-cycle, $c$ is an even permutation, hence $h$ is even, both as an element of $S_n$ and as an element of $S_{n-1}$.
Hence by the inductive hypothesis, regarding $h$ as an element of $S_{n-1}$, $h$ can be multiplied by a product of $3$-cycles $t_1,...,t_m$ in $S_{n-1}$ to reach the identity in $S_{n-1}$. Recasting $t_1,...,t_m$ as elements of $F_{n-1}$, it follows that $g{\cdot}c{\cdot}t_1\cdots t_m=e$ in $S_n$, which completes the induction.
This completes the proof.