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In-the-symmetric-group-S-n-o-let-H-denotes-the-set-of-permutations-leaving-the-integer-n-fixed-H-f-S-n-f-n-n-Show-that-the-pair-H-o-is-subgroup-of-S-n-o-note-the-operatio




Question Number 76303 by arkanmath7@gmail.com last updated on 26/Dec/19
In the symmetric group (S_n  , o), let  H denotes the set of permutations   leaving the integer n fixed:     H = {f∈S_n  ∣ f(n) = n}  Show that the pair (H, o) is subgroup  of (S_n ,o).  ′note: the operation is composition′
$${In}\:{the}\:{symmetric}\:{group}\:\left({S}_{{n}} \:,\:{o}\right),\:{let} \\ $$$${H}\:{denotes}\:{the}\:{set}\:{of}\:{permutations}\: \\ $$$${leaving}\:{the}\:{integer}\:{n}\:{fixed}: \\ $$$$\:\:\:{H}\:=\:\left\{{f}\in{S}_{{n}} \:\mid\:{f}\left({n}\right)\:=\:{n}\right\} \\ $$$${Show}\:{that}\:{the}\:{pair}\:\left({H},\:{o}\right)\:{is}\:{subgroup} \\ $$$${of}\:\left({S}_{{n}} ,{o}\right). \\ $$$$'{note}:\:{the}\:{operation}\:{is}\:{composition}' \\ $$
Answered by mind is power last updated on 26/Dec/19
Id∈H  let f ,g∈H  fog∈H since  fog(n)=f(g(n))=f(n)=n  if f∈H   f^− ∈H  since f∈S_n ⇒f is bijection⇒f^− existe un  f(n)=n⇒f^− of(n)=f^− (n)⇒n=f^− (n)  ⇒H is a sub groupe of (S_n_,  ,o)
$$\mathrm{Id}\in\mathrm{H} \\ $$$$\mathrm{let}\:\mathrm{f}\:,\mathrm{g}\in\mathrm{H} \\ $$$$\mathrm{fog}\in\mathrm{H}\:\mathrm{since} \\ $$$$\mathrm{fog}\left(\mathrm{n}\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{n}\right)\right)=\mathrm{f}\left(\mathrm{n}\right)=\mathrm{n} \\ $$$$\mathrm{if}\:\mathrm{f}\in\mathrm{H}\:\:\:\mathrm{f}^{−} \in\mathrm{H} \\ $$$$\mathrm{since}\:\mathrm{f}\in\mathrm{S}_{\mathrm{n}} \Rightarrow\mathrm{f}\:\mathrm{is}\:\mathrm{bijection}\Rightarrow\mathrm{f}^{−} \mathrm{existe}\:\mathrm{un} \\ $$$$\mathrm{f}\left(\mathrm{n}\right)=\mathrm{n}\Rightarrow\mathrm{f}^{−} \mathrm{of}\left(\mathrm{n}\right)=\mathrm{f}^{−} \left(\mathrm{n}\right)\Rightarrow\mathrm{n}=\mathrm{f}^{−} \left(\mathrm{n}\right) \\ $$$$\Rightarrow\mathrm{H}\:\mathrm{is}\:\mathrm{a}\:\mathrm{sub}\:\mathrm{groupe}\:\mathrm{of}\:\left(\mathrm{S}_{\mathrm{n}_{,} } ,\mathrm{o}\right) \\ $$$$ \\ $$
Commented by arkanmath7@gmail.com last updated on 26/Dec/19
Thanks
$${Thanks} \\ $$

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