Abstract algebra question (group theory)

[Check]

Been looking at this for a couple of days, can't figure out how to prove it. Any ideas?

Let G be a finite group with no subroups other than {e} or G itself. prove that G is either the trivial group {e} or a cyclic group of prime order.

 

[BMcDurl]

You don't even need the assumption that G is finite. Let a /in G, s.t. a!=e. By assumption the subgroup generated by a is G itself. Suppose G isn't finite. Then let b = a^2. b generates a proper subgroup of G. Contradiction. So we know G is finite. Now let |G| = n. Suppose n isn't prime. Let n = p * q, p > 1, q > 1, then consider b = a^p. So b^q = (a^p)^q = a^(pq) = a^n = e. Thus b generates a proper subgroup of G of order q. Contradiction.

 

[FallenHero]

The above post is why I never went into math.

 

[TuxDave]

Originally posted by: BMcDurl
You don't even need the assumption that G is finite. Let a /in G, s.t. a!=e. By assumption the subgroup generated by a is G itself. Suppose G isn't finite. Then let b = a^2. b generates a proper subgroup of G. Contradiction. So we know G is finite. Now let |G| = n. Suppose n isn't prime. Let n = p * q, p > 1, q > 1, then consider b = a^p. So b^q = (a^p)^q = a^(pq) = a^n = e. Thus b generates a proper subgroup of G of order q. Contradiction.

:Q So many words... and none of them make sense to me.

 

[Born2bwire]

Originally posted by: TuxDave

Originally posted by: BMcDurl
You don't even need the assumption that G is finite. Let a /in G, s.t. a!=e. By assumption the subgroup generated by a is G itself. Suppose G isn't finite. Then let b = a^2. b generates a proper subgroup of G. Contradiction. So we know G is finite. Now let |G| = n. Suppose n isn't prime. Let n = p * q, p > 1, q > 1, then consider b = a^p. So b^q = (a^p)^q = a^(pq) = a^n = e. Thus b generates a proper subgroup of G of order q. Contradiction.

:Q So many words... and none of them make sense to me.

"It's like he's trying to speak to me, I know it."

 

[jersiq]

Originally posted by: Born2bwire

Originally posted by: TuxDave

Originally posted by: BMcDurl
You don't even need the assumption that G is finite. Let a /in G, s.t. a!=e. By assumption the subgroup generated by a is G itself. Suppose G isn't finite. Then let b = a^2. b generates a proper subgroup of G. Contradiction. So we know G is finite. Now let |G| = n. Suppose n isn't prime. Let n = p * q, p > 1, q > 1, then consider b = a^p. So b^q = (a^p)^q = a^(pq) = a^n = e. Thus b generates a proper subgroup of G of order q. Contradiction.

:Q So many words... and none of them make sense to me.

"It's like he's trying to speak to me, I know it."

Little steps before the big steps. It's not as if this all popped into his head at once.
And if it did, then he's a frickin' genius :laugh:

 

[Stiganator]

"No--the Colonel--Is he here?" "I say you he be dead...."

 

[firewolfsm]

Actually, he's wrong. Don't be thrown off by all his variables, his post leads us nowhere.

 

[CP5670]

No, the argument is perfectly correct, just written in a condensed way. Although I don't know why he registered just to post that.

 

[hypn0tik]

Originally posted by: CP5670
No, the argument is perfectly correct, just written in a condensed way. Although I don't know why he registered just to post that.

I don't think the OP is going to complain about that, lol.

 

[Mday]

Man, its been so long since I've done this lol...

 

[Nathelion]

Brruah I finished that class this fall, never wanna look back...

 

[Mday]

Originally posted by: Nathelion
Brruah I finished that class this fall, never wanna look back...

I enjoyed my brush with modern algebra. Group, field, ring, etc.

 

[CP5670]

I'm a math student and did it a few years ago, but I'm more of an analysis guy and don't remember this stuff that well.

 

[rocadelpunk]

i miss working with groups. just one operation instead of two with rings : P

 

[Nathelion]

Originally posted by: Mday

Originally posted by: Nathelion
Brruah I finished that class this fall, never wanna look back...

I enjoyed my brush with modern algebra. Group, field, ring, etc.

15-20 hrs of homework per week contributed to my aversion for it.