Abelian Simple Group / Prime Numbers

[Rederick]

Homework Statement

If G is an abelian simple group then G is isomorphic to Zp for some prime p (do not assume G is a finite group).

Homework Equations

In class, we were told an example of a simple group is a cyclic group of prime order.

The Attempt at a Solution

Let G be an abelian simple group. Let a be an element in G that is not the identity. Let H be a cyclic subgroup of G, H=<a>. Since a is an element in G, it's non-trivial. Since G is abelian, H is normal in G. Since G is a simple group, it has only 2 normal subgroups, the identity and itself. Thus, H=G. Since G is cyclic and an abelian simple group, it has prime order. Hence G is isomorphic to Zp.

I feel uneasy about the last 2 sentences. Any advice to make this a more solid proof?

 

[ystael]

You have shown that whenever $$1 \neq a \in G$$, then $$\langle a\rangle = G$$. You need one more sentence to explain why this fact shows that $$|G|$$ is prime. (This sentence should replace "Since G is cyclic and an abelian simple group, it has prime order", which -- though you've added the observation that $$G$$ is cyclic -- is what you're trying to prove.)

 

[micromass]

Yes, the last 2 sentences are a bit to fast. So far you've only shown that a simple abelian group must be cyclic. You must still show now that cyclic simple groups are of prime order.

How do we show that? Take a cyclic group that is not of prime order. Try to find a proper subgroup in there. This shows that our group is not simple...

 

[stormyweathers]

an old thread, but I stumbled upon it looking for homework help so I figured I'd contribute anyway.

You must find a normal subgroup to show its not simple.