# Order of a Group with a Unique Subgroup

• Oct 18th 2010, 03:06 PM
matt.qmar
Order of a Group with a Unique Subgroup
Hello!

I have a problem I'm working on..
Given that a finite group G has exactly one nontrivial proper subgroup (call it H),

then we must prove
a) G must be cyclic
b) the order of G is \$\displaystyle p^2\$ for a prime p.

I have shown a) by noting that any element in G either generates
Case 1:the nontrivial proper subgroup H
Case 2:the group itself
So there is at least one element which generates the group (so it is cyclic).

I am having trouble showing part b), however.

Any help would be fantastic! and cause me to be very grateful! Thanks!
Cheers.
• Oct 18th 2010, 07:53 PM
tonio
Quote:

Originally Posted by matt.qmar
Hello!

I have a problem I'm working on..
Given that a finite group G has exactly one nontrivial proper subgroup (call it H),

then we must prove
a) G must be cyclic
b) the order of G is \$\displaystyle p^2\$ for a prime p.

I have shown a) by noting that any element in G either generates
Case 1:the nontrivial proper subgroup H
Case 2:the group itself
So there is at least one element which generates the group (so it is cyclic).

I am having trouble showing part b), however.

Any help would be fantastic! and cause me to be very grateful! Thanks!
Cheers.

So it is cyclic: it there were two primes dividing the order of the group then either Cauchy's Teorem or Sylow's theorems

would give us at least two different proper non-trivial subgroups, each of order a different prime, so the order of the

group is divisible only by one prime, say p. It can't be that the order is p, since then there is no proper non-trivial sbgp's at

all, and it can't be \$\displaystyle p^n\,,\,n\geq 3\$ , since then there are at least 2 different proper non-trivial sbgp's, of

orders\$\displaystyle p\,,\,p^2\$ (why??) , so...

Tonio