**To Prove** :

The order of every element of a finite group is finite and is less than or equal to the order of the group.

**Proof **:

Suppose G is a finite group, the composition being denoted multiplicatively. Suppose a ∈ G, consider all positive integral powers of a i.e., a, a^{2}, a^{3}, ……

All these are elements of G, by closure axiom.

Since G has a finite number of elements, therefore all these integral powers of **a **cannot be distinct elements of G.

Suppose,

a^{r}= a^{s}where r > s

Now

a^{r}= a^{s}=> a^{r}. a^{-s}= a^{s }. a^{-s}(multiplying both sides by a^{-s}) => a^{r}. a^{-s}= a^{0}( a^{s-s}= a^{0}) => a^{r}. a^{-s}= e => a^{m}= e, where m = r - s

Since

r > s

Therefore, ‘m’ is a positive integer. Hence, there exists a positive integer m such that a^{m} = e.

Now we know that every set of positive integers has the least member.

Therefore, the set of all those positive integers m such that a^{m} = e has the least members, say n. Thus, there exists the least positive integer n such that

a^{n}= e.

Therefore, the order of a, o(a) is finite.

Now to prove that o(a) ≤ o(G).

Suppose,

o(a) = n, where n > o(G).

Since a ∈ G, therefore by closure property a, a^{2}, …. a^{n} are elements of G. No two of these are equal. For if possible, let a^{r} = a^{s}, 1 ≤ s < r ≤ n. Then,

a^{r-s }= e

Since

0 < r - s < n

Therefore,

a^{r-s}= e implies that the order of a is less than n.

This is a contradiction. Hence, a, a^{2},… a^{n} are n distinct elements of G. Since n > o(G), therefore this is not possible.

Hence, we must have o(a) ≤ o(G).

Therefore, it is proved that The order of every element (a, o(a)) of a finite group (G) is finite and is less than or equal to the order of the group(i.e. o(a) ≤ o(g) ).