Let be a subset of the group
. Show that
is a subgroup of
if and only if
and
whenever
.
Attempts.
(only-if part) If is a subgroup of
, then
must contain the identity
and both
and
is inside
whenever
. Thus
and for any
,
is inside
. Thus
.
(if-part) First, because , there exists some
in
such that
(by the condition
whenever
). This proves the existence of the identity. Secondly, having proven
is in
, and noted that
whenever
, one proves the existence of the inverse. Thirdly,
implies
because
and also
.
is therefore a subgroup of
.
