**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 g^{n} 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 a^{s }∈ H, then the inverse of a^{s} 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

a^{m}∈ H

Then we shall prove that :

H = { a^{m }}

i.e., H is cyclic and is generated by a^{m} .

Let a^{t} be any arbitrary element of H.

By division algorithm, there exists integers q and r, such that :

t = mq + r, 0 ≤ r <m.

Now,

a^{m}∈ H ⇢(a^{m})^{q}∈ H ⇢ a^{mq}∈ H ⇢(a^{mq})^{-1}∈ H ⇢a^{-mq }∈ H.

Also,

a^{t}∈ H a^{-mq}∈ H ⇢ a^{t }a-^{mq}∈ H ⇢ a^{t-mq}∈ H ⇢ a^{r }∈ H. (Since, r = t- mq)

Now m is the least positive integer, such that :

a^{m}∈ H, 0 ≤ r <m.

Therefore, r must be equal to 0.

Hence,

t = mq

Therefore,

a^{t}= a^{mq }=(a^{m})^{q}.

Hence, every element a^{t }∈ H is of the form ( a^{m} )^{q} .

Therefore, H is cyclic and a^{m} 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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