**The order of an element of group:** Let a be a element of a group G. a is said to be of order m, if and only if m is the least positive integer such that a^{m} = e (the identity).

If there is no positive integer n such that a^{n} = e, a is said to be of infinite order or zero order.

The order of the element a is denoted by O(a).

**Torsion group**: A group G is called a torsion group if every element of G, is of finite order.

Example: The multiplicative group H = {1, -1, i, -i} and the additive group (Z_{8}; +) are torsion.

**Torsion free group**: A group G is called a torsion free group if no element in G, other than the identity, is of finite order.

Example: The additive group of integers Z is a torsion free group.

**Mixed group**: A group G is called a mixed group if and only if there exist at least two elements a, b ∊G such that

(i) a ≠ e and O(a) is finite.

(ii) O(b) = ∞.

**Cyclic group**: A group G is said to be a cyclic group if there exist an element

a ∊ G such that every element of G is expressible as an integral power of a i.e. as a^{k} where k is an integer.

## Theorem: Every cyclic group is Abelian group.

Proof: Let G be a cyclic group generated by a.

Let x , y be any two elements in G, then x, y are some integral powers of a. That is x = a^{r} and y = a^{s} where r and s some integers.

Now xy = a^{r }a^{s} = a^{r+s} = a^{s+r} = a^{s} a^{r} = yx

This implies that G is abelian.

## Theorem: Every finite cyclic group has exactly two generators.

Proof: Let G be an infinite cyclic group and a be a generator of G. Since the inverse of a generator is also a generator, a^{-1} must be a generator of G. We are to show that a and a^{-1} are the only generators of G. Since G is finite , a and a^{-1} are of finite order. This implies that for any two different integers r and s, a^{r} ≠ a^{s} and (a^{-1})^{r} ≠ (a^{-1})^{s}.

Let m be any integer different from 1 and -1. Then the element a^{m}∊ G such that a^{m} ≠ a^{1} and a^{m} ≠ a^{-1}.

Now for every integer k,

mk ≠ 1, -1, i.e. (a^{m})^{k }≠ a and (a^{m})^{k }≠ a^{-1}.

Thus a^{m }fails to generate G.

Hence G has only two generators a and a^{-1}.