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Proving elements of a finite group is finite
  • Last Updated : 17 Mar, 2021

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, 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.

Suppose,

ar = as    where r > s

Now                       

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 

Since    



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). 

Suppose,

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

Since 

0 < r - s < n

Therefore, 

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) ).

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