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So I did this task "Let A4 be the alternating group on 4 letters, that is the subgroup of S4 consisting of even permutations. Find elements σ1, σ2, σ3 ∈ A4 such that σ1 has order 1, σ2 has order 2, and σ3 has order 3."

and got σ1 = e, σ2 = (1, 2)(3, 4) σ3 = (1, 2, 3)

I'm now going to prove that A4 is not abelian, but I am stuck at computing σ2σ3

I first took σ2σ3 =(1,2)(3,4) * (1,2,3) = (1,4,2) because 1 maps to 2 then maps to 1, 2 maps to 3 then maps to 4 and 3 maps to 1 then maps to 2. But in the solution says σ2σ3 =(1,2)(3,4)(1,2,3)=(2,4,3)

Can someone explain this to me

user1162295
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    Permutations are functions, so if $f$ is such a function we write $f(1)$ for the value of $f$ in $1$. Now if we compute $f(1)$ for $f$ being the transposition $(12)$ then $f(1)=(12)$ of $1=2$. Now let us compute:$$(12)(34)(123)\text{ of }1=(12)(34)\text{ of }2= (12)\text{ of }2=1\ .$$Your convention of multiplying permutations corresponds to writing $(1)f$ instead of $f(1)$, and doing "the same" from left to the right. Some books and computer programs are doing so... – dan_fulea Aug 01 '23 at 12:02
  • Im sorry, but I didn't quite understand, could you explain further? – user1162295 Aug 01 '23 at 12:10
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    Left-to-right or right-to-left is not the issue here. The op says themself that 1 is fixed but they put it in a 3-cycle anyway. The problem is they're mixing up cycle and 2-row notation. – Matthew Towers Aug 01 '23 at 12:31
  • so how do I find that is not abelian? – user1162295 Aug 01 '23 at 12:55
  • and thanks for the help! – user1162295 Aug 01 '23 at 12:55

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