# Prove cyclic subroups => cyclic group

• October 11th 2009, 05:08 PM
elninio
Prove cyclic subroups => cyclic group
Prove (or disprove) If G is a group in which every proper subgroup is cyclic, then G is cyclic.
• October 11th 2009, 06:05 PM
siclar
Think small! What groups do you know of order < 6?
• October 11th 2009, 06:38 PM
elninio
Could an example of $S_3$ be used as proof?

I'm not too comfortable with subgroups or cyclic groups yet, but from my understanding, the subgroups of $S_3$ are:

<(1)>= {(1)}
<(1,2)> = {(1),(1,2)}
<(1,3)> = {(1),(1,3)}
<(2,3)> = {(1),(2,3)}
<(1,2,3)> = {(1), (1,2,3), (1,3,2)} = <(1,3,2)>

Since no cyclic subgroup is equal to all of $S_3$it isnt cyclic. Or, no x in $S_3$ for which $S_3$=x
• October 11th 2009, 06:43 PM
siclar
Actually I was thinking even smaller, the Klein Four group. All proper subgroups are of order 2, and so must be cyclic, but all elements are of order 2 and so the group is not cyclic. This is of course generalizes, since for any prime $p$, $\mathbb{Z}_p\times \mathbb{Z}_p$ has only proper subgroups of order $p$ which must be cyclic!
• October 11th 2009, 07:10 PM
elninio
Great example!
On a side note, I may be mistaken but if a group is cyclic, then all of its subgroups must also be cyclic, correct? I still dont understand the exact reason why the reverse of this is NOT true.
• October 11th 2009, 07:17 PM
siclar
You are absolutely correct that a subgroup of a cyclic group is cyclic. However, there is also exactly one subgroup of each order, which is not true of the Klein Four group.
• October 11th 2009, 08:15 PM
aman_cc
Quote:

Originally Posted by siclar
You are absolutely correct that a subgroup of a cyclic group is cyclic. However, there is also exactly one subgroup of each order, which is not true of the Klein Four group.

Thanks siclar. I was thinking if we make elninio's condition more strict that every proper subgroup is cyclic AND there is atmost one subgroup of any order - will his claim that G is cyclic hold true?
• October 11th 2009, 08:36 PM
siclar
Definitely in the finite case. If there is a unique subgroup of each order, then each proper subgroup is normal.Then for finite groups, in particular all the Sylow subgroups will be normal and cyclic forcing G to be cyclic. For infinite groups however, it is definitely less clear.

EDIT: Ha, of course! Just consider the infinite direct sum $\bigoplus_{p\text{ prime}}\mathbb{Z}_p$, in the sense that we only allow a finite number of coordinates be nonzero. Then any subgroup would have to be of finite order, lest we obtain an element with an infinite number of nonzero coordinates. Moreover it would have to "live" on a finite number of the $\mathbb{Z}_p$, and so would be a subgroup of the product of those groups which is cyclic, thus would have to be cyclic, yet the group is clearly not cyclic!