Question

 Prove that the product of an even permutation and an odd permutation is odd.

Answer 1

$Solution: Now, ~Our~ first ~claim~ is, ~every ~permutation~ can ~be ~expressed ~as ~\\the ~product ~of ~one ~and~ only~ one~ of~ the ~following: \\~~~~~~~~~~~~~~~~~~~~an~ odd ~number ~of ~transpositions    ⟺ odd~ permutation \\~~~~~~~~~~~~~~~~~~~~an~ even ~number ~of ~transpositions   ⟺ even~ permutation$

$There ~are ~many ~ways~ to ~write~ a~ permutation~ as ~the ~product~ of~ transpositions, \\~and ~they~ can ~vary~ in ~length,~ but~ those~ products~ will ~have~ either ~an~ odd~ or ~an \\~even~ number~ of~ factors, ~never~ both. \\We ~know ~that~ the ~number ~of~ transpositions~ in~ a~ product ~corresponding ~to~ a \\~permutation ~that~ is ~a ~cycle ~of ~length ~n ~can ~be ~expressed ~as ~the~ product ~of\\ (n−1)~ transpositions.$

One ~can~ always ~resort ~to~ following ~the~ pattern:~ Without ~loss~ of ~generality~ \\suppose ~n ~is ~odd ~then~\\~~~~~~~~~~~~~~~~~~~~~~~~~~~(a_1,a_2,...,a_n)=(a_1,a_n)(a_1,a_{n-1})...(a_1,a_2)\\which~is~even~because ~there ~are~ (n-1)~ transpositions.~\\ [we~ can~ also ~write ~ (a_1,a_2,...,a_n)=(a_1,a_2)(a_2,a_3)...(a_{n-1},a_n)]

$Again, ~an ~even ~number~ of~ transpositions   ⟺ the~ permutation~ is~ even.~\\ A~ similar~approach~ can ~be~ drawn~ for~ even~ number~ n.$

$So~ a ~cycle ~with~ a ~length~ that~ is~ even~ (has~ an ~even~ number~ of~ elements) ~is~ ODD, \\and~ a~ cycle~ with~ a~ length~ that ~is ~odd~ (has~ an ~odd~ number ~of ~elements)~ is~ EVEN.$ $If ~you~ have~ a~ permutation ~that~ is~ the ~product~ of~ disjoint ~cycles: ~say~ three~ cycles, ~\\corresponding ~to ~lengths~n_1,n_2,n_3,~then~ the~ number~ of ~transpositions~ \\representing~ this~ permutation ~can ~be~ computed ~by~ the ~parity~ of ~\\(n_1-1)+(n_2-1)+(n_3-1)~or ~simply~ the ~parity~(oddness/evenness) ~of~\\n_1+n_2+n_3-1.$ $Now ~Let ~E ~be~ the~ set ~of ~even~ permutations~ in~ G~ (which~ is~ presumably ~a ~group ~of ~\\permutations). Our ~second~ aim~ is ~to ~show~ that ~E ~forms~ a ~subgroup~ of ~G.$ $Let ~p~ and~ q~ be ~elements~ of~ E.~ Now~ we~ need ~to~ check ~if ~pq^{-1}~ is~ also~ an~ element ~\\of ~E. We~ know ~that ~a~ permutation ~is ~called~ an ~even ~permutation~ if ~it ~can ~be~\\ written ~as ~a~ product~ of~ an~ even~ number ~of~ transpositions.$

$So, let ~p=p_1p_2...p_{2k}~and~q=q_1q_2...q_{2j} ~ be ~a~ representation ~of~ p~ and ~q ~as ~a ~\\product ~of~ transpositions. \\We ~have~ that~q^{-1}=q_{2j}q_{2j-1}...q_2q_1~since ~transpositions~ are ~self ~inverses.$

$Thus, pq^{−1}=p_1p_2⋯p_{2k}q_{2j}⋯q_2q_1 ~is ~indeed~ a ~product ~of ~an ~even ~number ~of ~\\transpositions. Furthermore,~ pq^{−1}~is~ an ~element~ of ~G~ since~ p~ and ~q~ (and ~thus~ q^{−1})\\~ are ~elements~ of~ G~ and~ G ~is ~closed ~under~ products ~and ~inverses.$

$Thus, pq^{-1} \in E, implying~ that ~the~ identity ~is~ an~ element ~of~ E~ (by~ taking ~p=q), \\that~ it~ is~ closed~ under~ inverse ~(by~ taking~ p=id),~ and ~that ~it ~is~ closed~ under~\\ products ~(by ~taking ~q^{−1}~instead ~of q) ~and~ E~ is ~a~ subgroup ~of ~G.$

$Now~ suppose, ~if ~possible, ~there ~exist ~an ~even ~permutation ~p ~and~ an~ odd ~\\permutation q~ such ~that~ there~ product~r=pq~is~even~that~is~r \in E.~Then~\\q=p^{-1}r \in E~since ~p,r \in E~and ~E ~is~a~subgroup~of~G.Therefore, q ~is ~a ~even~ \\permutation~ but ~by~ assumption ~q ~is ~an~ odd~ permutation, ~contradiction.\\Therefore, ~product ~of~ an~ even~ permutation~ and~ an~ odd~ permutation~ is~ an~ odd ~\\permutation.$