# sub group of cyclic group

• Oct 17th 2009, 09:27 AM
aman_cc
sub group of cyclic group
Prove any subgroup (H) of a cyclic group (G) is itself cyclic

Approach:
Let $G = (a)$
Assume H is non-trivial
There is a $a^k$ such that k is the smallest +ve power among all $a^i$ 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 $H = (a^k)$
Let $a^i \in H$
$i = mk+r$ where $r
obviously $a^r$ $\in H$ (closure property of H)
Thus r can't be a +ve number (as k is the smallest). Thus r=0.
Hence $a^i = (a^k)^m$
And, $H = (a^k)$ (By definition)

Is this proof 'rigorous' enough?
• Oct 17th 2009, 09:52 AM
tonio
Quote:

Originally Posted by aman_cc
Prove any subgroup (H) of a cyclic group (G) is itself cyclic

Approach:
Let $G = (a)$
Assume H is non-trivial
There is a $a^k$ such that k is the smallest +ve power among all $a^i$ 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 $H = (a^k)$
Let $a^i \in H$
$i = mk+r$ where $r
obviously $a^r$ $\in H$ (closure property of H)
Thus r can't be a +ve number (as k is the smallest). Thus r=0.
Hence $a^i = (a^k)^m$
And, $H = (a^k)$ (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
• Oct 17th 2009, 10:09 AM
aman_cc
Quote:

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.