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?

I don't think $(ijk)$ multiplied by $(jki)$ is the identity permutation — J. W. Tanner, May 04 '20 at 10:36
@J.W.Tanner It is to fix those indices and all those are disjoint ,so they those indices $i$,$j$,$k$ will be fixed — Sandeep, May 04 '20 at 10:42
I'm guessing you have a typo. Multiplying $\ (ijk)\ $ by $\ (jki)\ $ doesn't fix anything. What you need to multiply by is $\ (kji)\ $: $$ (ijk)(jki)=(ikj)\ (ijk)(kji)=\text{identity} $$ — lonza leggiera, May 04 '20 at 10:49
Math mode isn't intended for italicizing text; as you can see here, it yields the wrong spacing for that. Usually text is italicized by enclosing it in asterisks. If you want to italicize text within math mode, you can use \mathit{...}- — joriki, May 04 '20 at 11:10
@quasi before lockdown was declared my professor was teaching cycles and alternating group but in middle lockdown was declared and it was in one of his lecture ,but he missed for the general case and just gave us basics and said us to learn as much as we can in holidays and it was bothering me from many days .So, i posted it — Sandeep, May 04 '20 at 11:46
@quasi no, but i inferred it.if it is wrong .could you say where i did wrong. — Sandeep, May 04 '20 at 11:53
@Sandeep: I think I misread the question. Ignore my previous comment (I misread it as "a product of $3$ cycles" rather than "a product of $3$-cycles"). — quasi, May 04 '20 at 11:57
Don't you still have a typo. where you say "if it has a three−cycles $(ijk)$ then i would multiply it by $(jki) $"? — J. W. Tanner, May 04 '20 at 12:00
and https://math.stackexchange.com/questions/914338/the-alternating-group-is-generated-by-three-cycles — J. W. Tanner, May 04 '20 at 15:30

quasi · Answer 1 · 2020-05-07T01:30:59.290

0

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.

edited May 07 '20 at 01:30

answered May 04 '20 at 12:40

quasi

61,115

What is the claim? Is it that $A_n$ is generated by $3$-cycles? I thought "I need to multiply by $3$-cycle to reach identity" in OP meant "I'm looking for an inverse," but perhaps it meant "I need to find a product of $3$-cycles that matches the given permutation" -- OP is hard to understand – J. W. Tanner May 04 '20 at 12:49
@quasi my question was if one of the $c_i$ is 5-cycle or higher – Sandeep May 04 '20 at 12:56
The element being expressed as a product of $3$-cycles is $g$. – quasi May 04 '20 at 12:57
@J.W.Tanner you understood the question right – Sandeep May 04 '20 at 12:57
@Sandeep: I showed that every even permutation is a product of $3$-cycles. As a consequence, every even permutation can be multiplied by a product of $3$-cycles to reach the identity (which is how I interpret the problem at hand). – quasi May 04 '20 at 12:59
@quasi yes, i understood it but if a permutation. $p$=$(5,1,2,3,4)$,how to reach to identity just by multiplying it by a $3-cycle$,any number of multiplications. – Sandeep May 04 '20 at 13:02
Not just one $3$-cycle. My proof was non-constructive. It shows that every even permutation can reach the identity by multiplying it by some finite number of $3$-cycles, but it doesn't give an explicit formula. – quasi May 04 '20 at 13:02
@Sandeep: For a constructive proof, see Robert Wolfe's answer in the post referenced in J. W. Tanner's comment. – quasi May 04 '20 at 13:08
@quasi yes not just one 3-cycle,ok but i need explicit formula for example a 5-cycle (i,j,k,l,m,n) to make it identity by mutiplying any number of 3-cycles – Sandeep May 04 '20 at 13:10
@Sandeep: For the example you asked for, one way is $$(1,2,3,4,5)(1,2,3)(1,2,3)(1,5,4)=e$$ – quasi May 04 '20 at 13:25

score 0 · Answer 2 · answered May 04 '20 at 14:57

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).

quasi · Accepted Answer · 2020-05-04T15:52:31.827

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.

Alternating groups using three-cycles

3 Answers3