Skip to content
Related Articles
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.

Attention reader! Don’t stop learning now. Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up