Finite Simple Groups: Exploring Order and Lagrange's Theorem

[ehrenfest]

[SOLVED] simple groups

Homework Statement

T or F: All nontrivial finite simple groups have prime order.

Homework Equations

The Attempt at a Solution

I want to say yes with Lagrange's Theorem, but I am not sure that applies...

 

[morphism]

It's false. A counterexample is going to be necessarily non-abelian (why?).

 

[ehrenfest]

I think I see. The reason is that the converse of Lagrange's theorem is false, right? I mean that if a group has order a and b divides a, there is not necessarily a subgroup of G that has an order of a.

The classic example is A_4. I believe this has no proper subgroups although it has order 12.

I am not exactly sure about the answer to your "why". Is it because any finite abelian group can be written as Z_p^r cross ... and you just multiply the p^r s to get the order of the group?

 

[morphism]

Yup - that the converse to Lagrange is false in general is basically why this statement is false. (Incidentally, the converse is true for abelian groups. Thus, the only finite simple abelian groups are those of prime order. And, in fact, any simple abelian group is finite.)

 

[Dick]

ehrenfest, you do know that 'simple group' means 'no normal subgroups', right? A_4 is a bad classic example. It has a both proper subgroups and a normal subgroup. I think what morphism is getting at is that ANY subgroup of an abelian group is normal. And Lagrange's theorem gives you one automatically if the order of the group is not prime.

 

[ehrenfest]

umm. so what is a counterexample then?

 

[morphism]

A very classical and popular one: A_5. This is the smallest finite simple non-abelian group.

 

[morphism]

I should expand a bit on my comments in #4. Suppose the converse to Lagrange's theorem held for all types of finite groups. Say we're given a nontrivial finite group G whose order is not prime. Let p be the smallest prime divisor of |G|. Then G must have (since we're assuming Lagrange's converse holds) a subgroup H of order |G|/p. This subgroup has index p in G, and is thus normal. (Here I'm using the following nifty fact: If H is a subgroup of a finite group G, and [G] is the smallest prime divisor of G, then H is normal.) Moreover, H is nontrivial and not the entire group. So G cannot be simple.

 

[Dick]

I suppose. But "supposing the converse to Lagrange's theorem held" is sort of ultra-hypothetical.

 

[morphism]

Definitely - and that was the point of my post - just hypothesizing about why a counterexample should exist. I guess I should have also explicitly mentioned that A_4 is not a counterexample to the statement in the OP, in post #4.

 

[Dick]

Too bad counterexamples do exist. Otherwise we could have gotten over this group classification problem before they invented computers. :)