Proof. Consider an element 1 in \(S\), there are \(n\) choices of \(\alpha (1)\). Once \(\alpha (1)\) is chosen, for \(\alpha (2)\), there are \(n-1\) choices, since \(\alpha \) is \(1-1\). Since \(\alpha (2)\) is chosen, there are \(n-2\) possibilities for \(\alpha (3)\). Continuing in this way, we see that \(S_n\) has \(n(n-1)(n-2)\cdots \cdots 3\cdot 2\cdot 1\, \) elements.

Thus \(\begin {vmatrix} S_n\\ \end {vmatrix} = n(n-1)(n-2)\cdots \cdots \, 3\cdot 2\cdot 1 = n!\)

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Proof. If \(\alpha \) and \(\beta \) are in \(S_n\) defined by \(\, \alpha = \begin {pmatrix} 1 & 2 & 3 & \cdots \cdots \\ 1 & 3 & 2 & \cdots \cdots \\ \end {pmatrix}\,\) and \(\, \beta = \begin {pmatrix} 1 & 2 & 3 & \cdots \cdots \\ 3 & 2 & 1 & \cdots \cdots \\ \end {pmatrix}\,\) with each element after 3 mapped onto itself. Then \(\, \beta o\alpha = \begin {pmatrix} 1 & 2 & 3 & \cdots \cdots \\ 3 & 1 & 2 & \cdots \cdots \\ \end {pmatrix}\,\) and \(\, \alpha o\beta = \begin {pmatrix} 1 & 2 & 3 & \cdots \cdots \\ 2 & 3 & 1 & \cdots \cdots \\ \end {pmatrix}\). Thus \(\alpha o\beta \neq \beta o\alpha \, \) and the group is non-Abelian.


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Proof. We can assume set \(S\) is of the form \(\, S = \{a_1a_2\cdots \cdots \, a_mb_1b_2\cdots \cdots \, b_nc_1c_2\cdots \cdots \, c_k\}\) where the \(c_{'s}\) are those elements which are fixed by both \(\alpha \) and \(\beta \). We want to show that for any permutation \(X\), \(\,\, (\alpha \beta )(X) = (\beta \alpha )(X)\). Now, suppose \(X\) is any of the elements say \(a_i\). Then; \[(\alpha \beta )(a_i) = \alpha (\beta (a_i)) = \alpha (a_i) = a_{i+1}\] since \(\beta \) fixes all \(a\) elements. Similarly \(\, (\alpha \beta )(b_j) = \alpha (\beta (b_j)) = \alpha (b_{j+1}) = b_{j+1}\,\) since \(\alpha \) fixes all \(b_js\) and \(\, (\beta \alpha )(b_i) = \beta (\alpha (b_j)) = \beta (b_j) = b_{j+1} \, \implies \alpha \beta = \beta \alpha \).
Now, since \(\alpha \) and \(\beta \) fix the \(cs\hspace {0.3cm}(\alpha \beta )(c_i) = \alpha (\beta (c_i)) = \alpha (c_i) = c_i)\) and
\((\beta \alpha )(c_j) = \beta (\alpha (c_j)) = \beta (c_j) = c_j\). Thus \(\alpha \) and \(\beta \) commute.


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Proof. Let \(\alpha \) be a permutation on \(S = \{a_1\,, \, a_2\, \cdots \cdots \,, \, a_n\}\). To write \(\alpha \) in cycle form, we say
\(a_2 = \alpha (a_1)\,\,\,\,\, a_3 = \alpha (\alpha (a_1)) =\alpha ^2(a_1)\). Proceeding in this till we get back at \(a_1 = \alpha ^m(a_1)\) for some \(m\). Clearly such an \(m\) exists because \(a_1\, , \, \alpha (a_1)\, , \, \alpha ^2(a_1)\, \cdots \cdots \) is finite so eventually there must be a repetition say \(\alpha ^i(a_1) = \alpha ^j(a_1)\) for some \(, i\, ,\, j\) with \(i<j\). Then \(a_1 = \alpha ^m(a_1)\) where \(m = j - i\). We express this relationship among \(a_1\, , \, a_2\,, \cdots \cdots \, a_m.\hspace {0.3cm}\alpha = (a_1a_2\cdots \cdots \, a_m)....\)
In case the set \(S\) is not exhausted, we choose another element \(b_1\) of \(S\) not in the cycle \(\alpha \). We set \(b_2 = \alpha (b_1)\, , \, b_3 = \alpha ^2(b_1).....\) and so on till \(b_1 = \alpha ^k(b_1)\) for some \(k\).

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Proof. Consider the \(r\) cycle (cycle of length \(r\)). If then \(\beta \) is the identity.
e.g \(\, \, \begin {pmatrix} 1 & 2 & 3\\ 1 & 2 & 3\\ \end {pmatrix} = (2\,3)(3\,2) = 1\) an identity map.
So \(\beta = (1\,2)(1\,2)\) since \(\, (1\, 2)^{-1} = 12\).
If \(r\geq 2\), then \(\beta = (1\, 2\, 3\,\cdots \cdots \, r)\) which can be written a \(\beta = (1\,r)(1\,(r-1))\cdots \cdots (1\,2)\).

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Proof.

(i)

Multiplying a permutation by a transposition changes the number of inversions by an odd amount, so it reverses parity. Hence \(\operatorname {Sgn}(\alpha \zeta ) = -\operatorname {Sgn}(\alpha )\).

(ii)

Write \(\beta \) as a product of \(k\) transpositions, so \(\operatorname {Sgn}(\beta ) = (-1)^{k}\). Applying part (i) \(k\) times to \(\alpha \beta \), \[\operatorname {Sgn}(\alpha \beta ) = (-1)^{k}\operatorname {Sgn}(\alpha ) = \operatorname {Sgn}(\alpha )\operatorname {Sgn}(\beta ).\] The general product formula follows by induction on the number of factors. □

Proof. Recall that for every transposition \(\zeta \), \(\,\, Sgn(\zeta ) = -1\). Therefore suppose \(\, \alpha = \zeta _1\, , \, \zeta _2\, \cdots \cdots \, \zeta _q\) is a function of \(\alpha \) into transposition \(\, Sgn(\alpha ) = Sgn(\zeta _1)\,Sgn(\zeta _2)\,\cdots \cdots \, Sgn(\zeta _q) = (-1)^q\). Therefore
If \(\, (-1)^q = 1\), then \(q\) is even \(\implies \alpha \) is even
If \(\, (-1)q = -1\), then \(q\) is odd \(\implies \alpha \) is odd.

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Solution. The elements of \(S_4\) include;

Therefore, the elements of \(A_4\). \[A_4 = \{(1)\, ,\, (1\,2\,3)(4)\, ,\, (1\, 3\, 2)(4)\, , \, (1\, 3)(2\, 4)\, , \, (1\, 4\,2)(3)\, , \, (1)(2\, 4\, 3)\, , \, (1\,2)(3\,4)\,,\, (1\, 4\, 3)(2)\, , \, (1\, 2\, 4)(3)\] \[\,,\, (1\, 3\, 4)(2)\, ,\, (1\, 4)(2\,3)\}\]

\(\therefore \,\, A_4\) has 12 elements.

Proof. For each old permutation \(\alpha \) in \(S_n\) the permutation \((1\,2)\alpha \) is even. Thus there are atleast as many even permutations as there are odd.
For \(\beta \) an even permutation, \((1\,2)\beta \) is odd. So, there are atleast as many odd permutations as there are even ones.
Thus the number of even permutations is equal to the numbers of odd permutations. Now \(\begin {vmatrix} S_n\\ \end {vmatrix} = n! \implies \begin {vmatrix} A_n\\ \end {vmatrix} = \dfrac {1}{2}\, n!\) where \(\begin {vmatrix} S_n\\ \end {vmatrix} = \) length of \(S_n\) (number of elements in \(S_n\)).


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