Proving a Finite Group is Not Simple

In summary, the conversation is about proving that if a finite group G on a finite set X has an element g that does not fix any element in X, and the order of G does not divide m!, then G is not a simple group. The suggested approach is to consider the permutation representation of G on X and show that the order of G divides the order of the symmetric group S_X, leading to a contradiction.
  • #1
tom.young84
23
0
I found this problem, and I was wondering if I'm on the right approach.

Let G be a finite group on a finiste set X with m elelements. Suppose there exist a g[tex]\in[/tex]G and x[tex]\in[/tex]X such that gx not equal to x. Suppose the order of G does not divide m!. Prove that G is not simple.

Would it suffice to show that an isomorphism "f" exists from G to X? Then we just need to prove two cases about the Ker(f). We need to show that Ker(f) can't just be the identity because then it would be an infinite group being isomorphic to a finite group. If the Ker(f)=G, then some stuff. Sorry for the informality, I'm not actually sure what happens if Ker(f)=G.
 
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  • #2
tom.young84 said:
Would it suffice to show that an isomorphism "f" exists from G to X?
What would that mean? X isn't necessarily a group, just a set.

Instead try to consider the permutation representation,
[tex]\varphi : G \to S_X[/tex]
afforded by the group action. We are told that for some g, [itex]\varphi(g)\not= 1[/itex] which tells you [itex]\ker \varphi \not= G[/itex]. If [itex]\ker\varphi=1[/itex], then [itex]\varphi[/itex] is an embedding so what is the order of [itex]\varphi(G)[/itex] and how does it relate to [itex]|S_X|[/itex]? Use this to show |G| divides [itex]|S_X| = m![/itex] which is a contradiction.
 

Related to Proving a Finite Group is Not Simple

1. What is a finite group?

A finite group is a group that has a finite number of elements. This means that the group has a finite number of elements that can be combined using a defined operation to produce another element in the group.

2. What does it mean for a finite group to be simple?

A simple group is a group that does not have any non-trivial normal subgroups. This means that the group cannot be broken down into smaller groups that maintain the same structure.

3. How can I prove that a finite group is not simple?

To prove that a finite group is not simple, you can show that the group has at least one non-trivial normal subgroup. This can be done by finding a subgroup that is closed under the group operation and is also a proper subgroup of the original group.

4. What are some common methods for proving a finite group is not simple?

There are a few common methods for proving a finite group is not simple. These include finding a proper normal subgroup, showing that the group is isomorphic to a known non-simple group, or using the Sylow theorems to show the existence of a non-trivial normal subgroup.

5. Can a finite group be both simple and not simple?

No, a finite group cannot be both simple and not simple. A group is either simple or not simple, there is no in-between. If a group is simple, it does not have any non-trivial normal subgroups. If a group is not simple, it has at least one non-trivial normal subgroup.

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