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Prove that every subgroup of a cyclic group is cyclic
  • Last Updated : 05 Mar, 2021

To Prove :
Every subgroup of a cyclic group is cyclic.

Cyclic Group :
It is a group generated by a single element, and that element is called a generator of that cyclic group, or a cyclic group G is one in which every element is a power of a particular element g, in the group. That is, every element of G can be written as gn for some integer n for a multiplicative group, or ng for some integer n for an additive group. So, g is a generator of group G.

Proof :
Let us suppose that G is a cyclic group generated by a i.e. G = {a}.
If another group H is equal to G or H = {a}, then obviously H is cyclic.
So let H be a proper subgroup of G. Therefore, the elements of H will be the integral powers of a.
If as ∈ H, then the inverse of as i.e;

a-s ∈ H

Therefore, H contains elements that are positive as well as negative integral powers of a.
Now, let m be the least positive integer such that

am ∈ H

Then we shall prove that :



H = { am }

i.e., H is cyclic and is generated by am .
Let at be any arbitrary element of H.
By division algorithm, there exists integers q and r, such that :

t = mq + r,    0 ≤ r <m.

Now,

  am ∈ H 
⇢(am)q ∈ H 
⇢ amq ∈ H
⇢(amq)-1 ∈ H
⇢a-mq  ∈ H.

Also,                 

at  ∈ H
a-mq  ∈ H ⇢ at a-mq  ∈ H
⇢ at-mq  ∈ H
⇢ ar  ∈ H.      (Since, r = t- mq)

Now m is the least positive integer, such that :

am ∈ H,      0 ≤ r <m.

Therefore, r must be equal to 0.
Hence, 

 t = mq                    

Therefore,

at = amq =(am)q .

Hence, every element at ∈ H is of the form ( am )q .
Therefore, H is cyclic and am is a generate of H.
Hence, it is proved that every subgroup ( in this case H) of a cyclic group ( G ) is cyclic.  

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