# Thread: sub group of cyclic group

1. ## sub group of cyclic group

Prove any subgroup (H) of a cyclic group (G) is itself cyclic

Approach:
Let $\displaystyle G = (a)$
Assume H is non-trivial
There is a $\displaystyle a^k$ such that k is the smallest +ve power among all $\displaystyle a^i$ that are in H. Well order principle of natural numbers ensures there will always be a unique such k as long as H is non-trivial
Claim $\displaystyle H = (a^k)$
Let $\displaystyle a^i \in H$
$\displaystyle i = mk+r$ where $\displaystyle r<k$
obviously $\displaystyle a^r$$\displaystyle \in H (closure property of H) Thus r can't be a +ve number (as k is the smallest). Thus r=0. Hence \displaystyle a^i = (a^k)^m And, \displaystyle H = (a^k) (By definition) Is this proof 'rigorous' enough? 2. Originally Posted by aman_cc Prove any subgroup (H) of a cyclic group (G) is itself cyclic Approach: Let \displaystyle G = (a) Assume H is non-trivial There is a \displaystyle a^k such that k is the smallest +ve power among all \displaystyle a^i that are in H. Well order principle of natural numbers ensures there will always be a unique such k as long as H is non-trivial Claim \displaystyle H = (a^k) Let \displaystyle a^i \in H \displaystyle i = mk+r where \displaystyle r<k obviously \displaystyle a^r$$\displaystyle \in H$ (closure property of H)
Thus r can't be a +ve number (as k is the smallest). Thus r=0.
Hence $\displaystyle a^i = (a^k)^m$
And, $\displaystyle H = (a^k)$ (By definition)

Is this proof 'rigorous' enough?

Rigorous and nice enough. Mention though something about dividing an integer by another integer, euclidean property or something like that.

Tonio

3. Originally Posted by tonio
Rigorous and nice enough. Mention though something about dividing an integer by another integer, euclidean property or something like that.

Tonio

Thanks Tonio. Will do that.