Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 4 / 2 / ii

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- Solution ii

Solution

Let

x, y \in \overline{H}

and let

W

be an open neighbourhood of

x y^{- 1}

. Continuity of

(u, v) \mapsto u v^{- 1}

supplies neighbourhoods

U

of

x

and

V

of

y

with

U V^{- 1} \subseteq W

. Choose

h \in H \cap U

and

k \in H \cap V

. Then

h k^{- 1} \in H \cap W

, so

x y^{- 1} \in \overline{H}

. The subgroup criterion using

x y^{- 1}

proves the closure of a subgroup is a subgroup. The neighbourhood argument avoids assuming the topology is determined by sequences.

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