Every subgroup of a cyclic group
is cyclic. If
generates the group
(so that
) and the order of
is
so that
then the order of any subgroup of
is a divisor of
and for each divisor
of
the group
has exactly one subgroup of order
– the subgroup generated by![]()
Proof: Let
and suppose that
is a subgroup of
If
consists of the identity element then
is cyclic, so assume
and
contains some element![]()
then
Let
be the least positive integer such that
The claim is that
Let
be an arbitrary member of
Since
we have
for some
Apply the division algorithm to
and
to obtain a quotient
and remainder
satisfying
so that
but
and
hence
but
is the least power of
in
and since
we must have
so that![]()
Conversely suppose that
is any subgroup of
We have already shown that
for some
Since
we have that the order of
is a divisor of
hence the order of![]()
is a divisor of![]()
Now let
be any divisor of
and let![]()
so no smaller power of
equals
so
has order![]()
is the only subgroup of order
since if
is any other subgroup of order
then
where
is the least positive integer such that
Write
with
We have
so that
Then
and
Also
so
and![]()