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.
Suppose G is a finite group, the composition being denoted multiplicatively. Suppose a ∈ G, consider all positive integral powers of a i.e., a, a2, a3, ……
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.
ar = as where r > s
ar = as => ar . a-s = as . a-s (multiplying both sides by a-s ) => ar . a-s = a0 ( as-s = a0) => ar . a-s = e => am = e, where m = r - s
r > s
Therefore, ‘m’ is a positive integer. Hence, there exists a positive integer m such that am = 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 am = e has the least members, say n. Thus, there exists the least positive integer n such that
an = e.
Therefore, the order of a, o(a) is finite.
Now to prove that o(a) ≤ o(G).
o(a) = n, where n > o(G).
Since a ∈ G, therefore by closure property a, a2, …. an are elements of G. No two of these are equal. For if possible, let ar = as, 1 ≤ s < r ≤ n. Then,
ar-s = e
0 < r - s < n
ar-s = e implies that the order of a is less than n.
This is a contradiction. Hence, a, a2,… an 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) ).