Alternating groups using three-cycles

Question

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?

Answer 1

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.

Answer 2

Of course we need $n > 2$.

Assume $n\ge 3$.

Our goal is to show that every element of $A_n$ can be multiplied by a product of $3$-cycles to reach the identity.

We start by proving the following claim . . .

Claim:$\;$Every element of $A_n$ is a product of $3$-cycles.

Proof:

For $3\le n\le 4$, the claim can be verified by direct computation (write each even permutation explicitly as a product of $3$-cycles).

Next assume $n\ge 5$.

Let $G$ be the subgroup of $S_n$ generated by the set of $3$-cycles of $S_n$.

Since $3$ is odd, all $3$-cycles are even permutations, hence $G$ is a subgroup of $A_n$.

Since $n\ge 5$, $A_n$ is simple.

Note that $G$ is not the trivial group.

Thus, to show that $G=A_n$ it suffices to show that $G$ is a normal subgroup of $A_n$.

Let $g\in G$ and $a\in A_n$.

Since $g\in G$ we can write $g=c_1\cdots c_m$ where each $c_i$ is a $3$-cycle.

Then $aga^{-1}=(ac_1a^{-1})\cdots (ac_ma^{-1})$, which is a product of $3$-cycles, since the conjugate of a $3$-cycle is a $3$-cycle.

Hence $aga^{-1}\in G$, which proves that $G$ is a normal subgroup of $A_n$.

Therefore $G=A_n$, which completes the proof of the claim.

Returning to the original goal, let $g\in A_n$. Then $g^{-1}$ is a product of $3$-cycles, hence from $gg^{-1}=e$, it follows that $g$ can be multiplied by a product of $3$-cycles to reach the identity.

Note:$\;$With the same reasoning, we get the following more general result . . .

Fix an odd positive integer $k\ge 3$, and let $n$ be any positive integer such that $n\ge k$.$\;$Then

• Every element of $A_n$ is a product of $k$-cycles.$\\[4pt]$
• Every element of $A_n$ can be multiplied by a product of $k$-cycles to reach the identity.

Answer 3

• You have already shown that a product $\ (ij)(kl)\ $ of two disjoint transpostions can be written as the product $\ (ijk)(jkl)\ $.
• The product of two transpositions which share a single entry can be written as $\ (ij)(jk)\ $ without loss of generality, and this is just the $3$-cycle $\ (ijk)\ $.
• The only other type of product of two transpositions is for the two to be identical, in which case their product is the identity.

Any even permutation can always be written as a product of an even number of transpositions in which successive pairs are either of the first or second of the above types. Each successive pair can thus be replaced either by a single $3$-cycle (if it's of the second type) or a product of two $3$-cycles (if it's of the first type).