 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-i/5/a/i/solution

f, g : X \to X

(f \circ g)^{- 1} = g^{- 1} \circ f^{- 1}

Sym (X)

. It is associative, since both

(f \circ g) \circ h

and

f \circ (g \circ h)

send

x

f (g (h (x)))

id_{X}

is a bijection and satisfies

f \circ id_{X} = id_{X} \circ f = f

f

f^{- 1} : X \to X

, itself a bijection, satisfying

f \circ f^{- 1} = f^{- 1} \circ f = id_{X}

. These verify all group axioms, so $Sym (X)$ is a group, the symmetric group of

X

 New to topics? Read the docs here!