If a group of order $\displaystyle p^m$, where p is an odd prime, contains a unique subgroup of order $\displaystyle p^s$, for some $\displaystyle 0 <s <m $,

then G must be cyclic.

How to prove this ?. Plz help.

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- Apr 23rd 2013, 09:39 AMthippliGroup of prime power order with unique subgroup
If a group of order $\displaystyle p^m$, where p is an odd prime, contains a unique subgroup of order $\displaystyle p^s$, for some $\displaystyle 0 <s <m $,

then G must be cyclic.

How to prove this ?. Plz help. - Apr 23rd 2013, 10:37 AMHallsofIvyRe: Group of prime power order with unique subgroup
Suppose it were NOT cyclic. Then there exist elements a and b such that a is not a power of b and b is not a power of a. Show that the set of powers of a form a subgroup and the set of powers of b form a different subgroup.

- Apr 24th 2013, 02:17 AMthippliRe: Group of prime power order with unique subgroup
but here the condition is "for some 0 < s < m " not "for all 0 < s < m ".

- Apr 24th 2013, 11:58 AMjohngRe: Group of prime power order with unique subgroup
Hi,

I've attached a solution. Initially, I was going to provide all the details, but then I realized I was in essence proving the theorem quoted first. So I just quoted the theorem with a reference.

Attachment 28124

Attachment 28125