# Questions involving simple groups

1. Jun 22, 2008

### cakesama

Hi, it's my first time posting in this forum, so I'm sorry if I have done anything against the forum rules and please point it out to me. Currently revising group theory for an exam in a week's time, and these two practise questions I couldn't finish, so if anyone can push me towards the right direction it'll be great.

The problem statement, all variables and given/known data

1. Prove that a simple abelian group G is cyclic of prime order.

2. Let G be a simple group and have a subgroup of index k > 2. Prove that |G| divides k!/2.

The attempt at a solution

Question 1
Suppose G is simple abelian, and consider an element x in G where x is not the identity element. Then since G is abelian, the cyclic group generated by x, <x> is normal in G, and this implies that G = <x>, so G is cyclic.

But now I'm not sure how to prove the fact that G is finite and has prime order...

Question 2
Suppose H is a subgroup of G, then by LaGrange |G| = |H||G| = |H|k. Also since k > 2, clearly k!/2 = k*(k-1)*(k-2)*...*3. So |G| divides k!/2 for any k > 2.

But I didn't even use the fact that G is simple, and this argument seems far too simple and I don't feel confident with the solution...

2. Jun 22, 2008

### CompuChip

For the first: Suppose it does not have prime order, then there is some d which divides |G|. Try to make a normal subgroup out of that (e.g. look at $x^{|G|/d}$). If G is not finite, then x has infinite order and G is not cyclic. For example, G = (Z, +) with x = 1 is such a group. Why doesn't the theorem work there?

As for the second, I easily see that k divides k!/2, but you didn't make it quite clear why |H| divides (k-1)*(k-2)*...*3.

3. Jun 22, 2008

### matt grime

No, infinite cyclic groups are allowed. But clearly all infinite cyclic groups are isomorphic to the integers, and this is the only exceptional case to prove.

4. Jun 22, 2008

### matt grime

For 2, let's examine the evidence: I need to show that something has order dividing k!/2. Let's focus on the k! first. As you've invoked Lagrange's theorem, you should be happy seeing that this implies that you need to do something with subgroups of groups of order k!. WHat group has order k!? Now, is there anyway we can naturally get this group occurring in this question?

5. Jun 22, 2008

### cakesama

Another attempt using the hints:

Question 1
following on from my first post,
Now for a contradiction, suppose that G does not have a prime order p.

case1 : G finite, |G|=n
Then G=<x>, and there exists a d that divides |G|, so by the fundamental theorem of cyclic groups there is a cyclic subgroup of G, <$x^{n/d}$>. And again G abelian implies that this subgroup is normal, so we have a proper normal subgroup of G. CONTRADICTION

case2 : G infinite,
G infinite cyclic so G is isomorphic to the integers under addition but the integers are not simple as 2Z is a nontrivial normal subgroup of the integers. So G is not infinite.

Therefore G is finite, and G has prime order.

Question 2

Still not totally sure on this one...
The symmetric group of degree k has order k! and the alternating group of degree k has order k!/2. Am I supposed to utilize the fact that the alternating group of degree k is simple for k is greater than or equal to 5?

6. Jun 22, 2008

### matt grime

That the alternating group is simple isn't important: that the symmetric group is *not* simple is, and any simple subgroup of S_n must be a subgroup of A_n.

Now, follow what you're given: G is simple, if I can show G is isomorphic to a subgroup of A_n you're done. What do you know about group homomorphisms from a simple group to any group?

7. Jun 22, 2008

### cakesama

I *think* I got it this time,

If we define a nontrivial homomorphism theta from G to A_k, then the kernel of theta must be trivial since G is simple, and theta is nontrivial. So then by the 1st isomorphism theorem, the mapping from G to G' (G' is the image of G under this mapping) is an isomorphism, and G' is a subgroup of A_k. Hence G is isomorphic to a subgroup of A_k, in particular they must have the same order. Therefore G divides the order of A_k.

What do you think?

8. Jun 23, 2008

### morphism

The keyword in that sentence is "if". You haven't shown that there is such a homomorphism.

9. Jun 24, 2008

### cakesama

Also, I was working on some other questions as well and came to another problem that I couldn't finish because I couldn't explicitly define a homomorphism!

If I can define a homomorphism, then things will start to work... could someone please tell me if it is possible to always define a nontrivial homomorphism between two arbitrary groups? The only thing I can think of is Cayley's theorem which isn't going to be useful in this case... I searched books in the library and the internet and got nowhere :(

10. Jun 24, 2008

### matt grime

No it clearly is not possible to do. And you've already written the facts in this post which preclude that: let G be simple, and K arbitrary with |G| not dividing |K|. Then the only homomorphism from G to K is the trivial one.

Getting back to the question. What information have you not used? Ans: G has a subgroup of index k. You're attempting to realize G as a subgroup of the permutations of a set of order k. How are you going to get a set of order k from the information that G has a subgroup of index k?