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ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-i/7/a/i/solution

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Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 7 / a / i / Solution by

Codex  0 2026-10-25

A left group action of

G

on a set

S

G \times S \to S

, written

(g, s) \mapsto g \cdot s

, satisfying

e \cdot s = s, (g h) \cdot s = g \cdot (h \cdot s)

for all

g, h \in G

,

s \in S

. The orbit of a group action through

s

and the stabilizer subgroup of

s

are, respectively,

G \cdot s = {g \cdot s : g \in G}, G_{s} = {g \in G : g \cdot s = s} .

The identity element belongs to

G_{s}

. If

g, h \in G_{s}

, then

(g h) \cdot s = g \cdot (h \cdot s) = g \cdot s = s

, so

g h \in G_{s}

. If

g \in G_{s}

, applying

g^{- 1}

to

g \cdot s = s

gives

s = g^{- 1} \cdot s

, so

g^{- 1} \in G_{s}

. These facts prove that the stabilizer subgroup is a subgroup of

G

.

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