 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2014/prelims/mathematics-i/7/a/ii/solution

ϕ_{g} \circ ϕ_{g^{- 1}} = ϕ_{e} = id_{S}, ϕ_{g^{- 1}} \circ ϕ_{g} = id_{S} .

Thus

ϕ_{g}

is a bijection with inverse

ϕ_{g^{- 1}}

S

. Also, for every

s

ϕ_{g h} (s) = ϕ (g, ϕ (h, s)) = (ϕ_{g} \circ ϕ_{h}) (s) .

Therefore the assignment

Φ (g) = ϕ_{g}

takes values in the symmetric group

Sym (S)

and satisfies

Φ (g h) = Φ (g) Φ (h) .

It is a group homomorphism, with the group product on the right interpreted as composition of permutations.

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