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So I started studying the symmetric group, and was given this statement:

Suppose that $(S_n, \circ)$ is a symmetric group with $n \geq 3$, then a permutation of $(i,j)=(1,i)(1,j)(1,i)$

And I fail to understand why is that the case. Even by drawing an example, I fail to see why that is the case. Could someone clarify this, please?

user401855
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    to be thourough you should stipulate that $1,i,j$ are different else the formula doesn't always work. It works if $i=1$ and $i\ne j$ for instance but not if $j=1$ and $i\neq j$. – zwim Sep 16 '17 at 20:47

2 Answers2

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Omiting all the fixed points in the second (bracket-) representation we get

$(1,i) = \begin{bmatrix} 1 & i & j \\ i & 1 & j \end{bmatrix}$

therefore

$(1,j) \circ (1,i) = \begin{bmatrix} 1 & i & j \\ i & j & 1 \end{bmatrix} = (1,i,j)$

and so we conclude

$(1,i) \circ (1,j) \circ (1,i) = \begin{bmatrix} 1 & i & j \\ 1 & j & i \end{bmatrix} = (i,j)$.

Does that make it clearer now?

EDIT: If you're not familiar with the bracket notation: It just shows which value on the top row gets mapped to which value on the bottom row. Furthermore I use the convention that if $f=(a,b,c)$ then $f(a) = b$, $f(b) = c$ and lastly if $a,b,c$ are premutations I interpret $abc = a \circ b \circ c$ so $(abc)(x) = a(b(c(x)))$.

flawr
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The only elements which are moved are $1,i,j$. Let's what happens to each in detail:

  • $1\mapsto i=i\mapsto 1$;
  • $i\mapsto 1=1\mapsto j =j$;
  • $j=j\mapsto1\mapsto i$.
Bernard
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