A non empty subset *H* of a group *G* is a *subgroup* of *G*
if, under the binary operation defined on *G*, *H* is a group.

In order to test whether a subset is a subgroup it is sufficient to complete the following three tests:

- Show
- Show if
*a*, then - Show if then

For example, the even integers are a subgroup of the integers under addition. Certainly the identity 0 is an even integer, the sum of two even integers is an even integer and the inverse of an even integer is an even integer.

The rotations and the identity in are a subgroup of . However, the reflections in together with the identity do not constitute a subgroup.

For any group *G* and , let be the
subset defined by . This is a subgroup of
*G* and is the cyclic subgroup generated by *x*.
So, in the non zero real numbers under multiplication the set
is a cyclic subgroup.

A very important result concerning subgroups of finite groups is
*Lagrange's Theorem* which states that the order of a
subgroup divides the order of a group.

With regard to finite cyclic groups, if *G* is a
finite cyclic group of order *n* and *x* is a generator, then
is a cyclic subgroup of order .
Indeed, every subgroup of a cyclic group is cyclic. Now Lagrange's
Theorem would tell us that a cyclic subgroup of a finite cyclic
group of order *n* must have order a divisor of *n*. In this case,
the converse is true, namely, for every divisor *m* of *n* there is
a cyclic subgroup of order *m*. It is generated by where
*x* is a generator of the cyclic group and *k* satisfies *km* = *n*.
This converse statement is not true for groups in general.

Sun Mar 30 14:48:35 PST 1997