Subgroups and cyclic groups

A subgroup $H$ of a group $G$ uses the operation already defined on $G$. Associativity is inherited, so there is no need to prove it again. What must be checked is that the identity lies in the subset and that products and inverses stay inside it. The empty subset fails immediately. A set of invertible matrices can also fail: diagonal matrices with positive entries are closed under products and inverses, whereas diagonal matrices whose entries exceed one do not contain the identity.

The standard subgroup test packages these checks: a nonempty subset is a subgroup if $xy^{-1}$ lies in it whenever $x$ and $y$ do. Taking equal elements supplies the identity. Taking the identity as $x$ supplies inverses. Applying the condition again with an inverse in the second slot supplies ordinary products. This proof explains why nonemptiness cannot be omitted: otherwise the stated condition would be vacuously true for an empty set.

For an element $g$, its cyclic subgroup consists of all integer powers, including negative powers and the zeroth power. It is the smallest subgroup containing $g$, because any subgroup containing $g$ must also contain its inverse and every repeated product. Integer powers obey $g^m g^n=g^{m+n}$. This exponent law is safe even in a nonabelian ambient group because both factors are powers of the same element.

The order of an element is the smallest positive exponent returning it to the identity, if such an exponent exists. If none exists, the element has infinite order. A finite order $n$ means the powers with exponents zero through $n-1$ are all distinct and every later power repeats one of them. Thus the element order equals the size of its cyclic subgroup. The order of the ambient group is a different quantity and can be larger.

Write the cyclic group of order $n$ additively as integers modulo $n$. Repeating the element represented by $a$ means taking multiples of $a$, with residues identified modulo $n$. It returns to zero when $n$ divides $ka$. Dividing both $n$ and $a$ by their greatest common divisor reduces this to the first return time $n/\gcd(n,a)$. The element generates the whole group exactly when that common divisor is one.

Subgroups of an infinite cyclic group are also cyclic. For a nonzero subgroup of the integers, choose its smallest positive element $d$. Divide any other member by $d$ with remainder. Closure under subtraction puts the remainder in the subgroup, and minimality forces it to be zero. Hence every member is a multiple of $d$. This least-positive-element argument also guides the classification of subgroups of a finite cyclic group; it is stronger than simply listing a few examples.