Prove Finite Simple Group has no Subgroup of Index 5

In summary, we have proven that if G is a finite simple group of order greater than 5 that contains no elements of order 2, then G contains no subgroup of index 5. This was done by showing that if such a subgroup H exists, then it must have an odd order and be isomorphic to a subgroup of S(5), which would contradict the simplicity of G. Therefore, G does not have any subgroups of index 5.
  • #1
bham10246
62
0
Question: Suppose that [itex]G[/itex] is a finite simple group of order greater than [itex]5[/itex] that contains no elements of order [itex]2[/itex]. Prove that [itex]G[/itex] contains no subgroup of index [itex]5[/itex].

Attempt: Suppose G has a subgroup H of index 5, i.e., [G:H]=5. Since 2 doesn't divide |G|, 2 doesn't divide |H|. So the order of H (and the order of G) is odd.

If |H|=1, then |G|=5 x |H|= 5. This is a contradiction because |G|>5. So |H|> 2.

Now if [itex]|G|=(5^a) m [/itex] where [itex]5 \not | m[/itex], then for [itex]n_5 = \#(Sylow\: 5\: subgroups)[/itex], [itex]n_5 | m [/itex] implies that [itex]n_5[/itex] is odd. If [itex]n_5 =1 [/itex], this contradicts that G is simple. So [itex]n_5 \geq 3[/itex].

From Sylow's theorem, we also know that [itex]n_5 \equiv 1 \mod 5[/itex]. So the last digit of [itex]n_5[/itex] must be 1.

Okay, I think this is the correct approach but I don't see any contradiction. Please help...

Thank you.
 
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  • #2
I thought that given a finite group G and any subgroup H of G, we have [G:H]=|G|/|H| according to Lagrange's Theorem. Maybe it's my mistake?

I just looked up Lagrange's Theorem and I think for any subgroup H, the index of H in G is defined as [G:H]=|G|/|H|.
 
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  • #3
The index of s subgroup is indeed just the number of cosets - there is no requirement for the subgroup to be normal at all.
 
  • #4
Kummer said:
A finite simple group is a finite group which has not non-trivial proper normal subgroup. So how it is possible for [tex](G:H)=5[/tex] if that expression is only defined for groups [tex]H[/tex] so that [tex]H\triangleleft G[/tex] unless [tex]H=\{ e \}[/tex]. But then that is trivial. So perhaps you mean to ask the group is a finite group.
What?

As for the original question, there is a theorem that states: If G is a finite simple group and H is a proper subgroup of G, then |G| divides [G:H]! (factorial).

This should help you out.

Another result that could be used is: If G is a finite group and p is the smallest prime divisor of |G|, then any subgroup of index p in G is normal.

Both results can be proved using group actions.
 
  • #5
bham10246 said:
Question: Suppose that [itex]G[/itex] is a finite simple group of order greater than [itex]5[/itex] that contains no elements of order [itex]2[/itex]. Prove that [itex]G[/itex] contains no subgroup of index [itex]5[/itex].

Attempt: Suppose G has a subgroup H of index 5, i.e., [G:H]=5. Since 2 doesn't divide |G|, 2 doesn't divide |H|. So the order of H (and the order of G) is odd.

If |H|=1, then |G|=5 x |H|= 5. This is a contradiction because |G|>5. So |H|> 2.

Now if [itex]|G|=(5^a) m [/itex] where [itex]5 \not | m[/itex], then for [itex]n_5 = \#(Sylow\: 5\: subgroups)[/itex], [itex]n_5 | m [/itex] implies that [itex]n_5[/itex] is odd. If [itex]n_5 =1 [/itex], this contradicts that G is simple. So [itex]n_5 \geq 3[/itex].

From Sylow's theorem, we also know that [itex]n_5 \equiv 1 \mod 5[/itex]. So the last digit of [itex]n_5[/itex] must be 1.

Okay, I think this is the correct approach but I don't see any contradiction. Please help...

Thank you.
@matt_grime. A mistake I made.

You got it. The last digit must be one. Now [tex]G=5^am[/tex] since [tex]n_5|G[/tex] it means [tex]n_5=5^bc[/tex] where [tex]c|m[/tex] and [tex]b\leq a[/tex]. But [tex]5^bc[/tex] never ends in 1.
 
  • #6
Kummer said:
You got it. The last digit must be one. Now [tex]G=5^am[/tex] since [tex]n_5|G[/tex] it means [tex]n_5=5^bc[/tex] where [tex]c|m[/tex] and [tex]b\leq a[/tex]. But [tex]5^bc[/tex] never ends in 1.
What if b=0 (which it must be, by Sylow)?
 
  • #7
morphism said:
What if b=0 (which it must be, by Sylow)?

Then we have,
[tex]c\equiv 1 (\bmod 5)[/tex] by Sylow's third theorem.
But [tex]c[/tex] is odd because "G has no element of order 2".


It seems I am missing something, very tired right now. But it makes sense to me now.

EDIT: Yes, a mistake. Because c=11, is not a contradiction.
 
  • #8
duhhh, now i remember a main point of the course i taught last year.:

if a group G has a sub group H of index n, then there is a non trivial map G-->S(n) given by the action of G on by translation, on the cosets of H.

hence if G has a subgroup of index 5, there is a non trivial homomorphism G-->S(5).

But if G is simple it is injective, then G has order less than 60, i.e. G is isomorphic to A(5), which has an element of order 2, contradiction to hypothesis.
 
  • #9
Yeah, that's basically the proof of the |G| divides [G:H]! thing.

Once you embed G in S(5), you can finish it off differently by concluding that |G|=15, and then:
1) S_5 doesn't have a subgroup of order 15. Contradiction.
2) G =~ C_15, which is not simple (because the only simple abelian groups are the prime cyclic ones), or
3) G has an element of order 5 by Cauchy, and thus a subgroup of index 3 which must be normal (by the second result I posted). So again G cannot be simple.
 
  • #10
Thanks for all your help! I was traveling for a few days but I'm back into the studying mode. Yes, you guys are right. You're absolutely right! :cool:
 

1. What is a Finite Simple Group?

A Finite Simple Group (FSG) is a group in abstract algebra that has no proper nontrivial normal subgroups. In other words, the only normal subgroups of an FSG are the trivial subgroup (containing only the identity element) and the entire group itself.

2. What is a subgroup of index 5?

A subgroup of index 5 is a subgroup that divides the size of the original group by 5. In other words, if a group has 100 elements, a subgroup of index 5 would have 20 elements.

3. Why is it important to prove that a Finite Simple Group has no subgroup of index 5?

Proving that a Finite Simple Group has no subgroup of index 5 is important because it helps to classify and understand the structure of these groups. It also has applications in other areas of mathematics and physics, such as in the study of finite group representations.

4. How can one prove that a Finite Simple Group has no subgroup of index 5?

There are several methods that can be used to prove that a Finite Simple Group has no subgroup of index 5. One approach is to use the fact that all FSGs have even order, and thus any subgroup of index 5 would have odd order, which is a contradiction. Another approach is to use the structure theorem for FSGs, which states that they can only be isomorphic to certain known groups with specific properties.

5. Are there any exceptions to the statement "Prove Finite Simple Group has no Subgroup of Index 5"?

Yes, there are a few exceptions to this statement. The most well-known exception is the group A5, which is the alternating group of degree 5. A5 is a finite simple group that does have a subgroup of index 5, namely the subgroup generated by a 5-cycle. However, this is the only exception and all other finite simple groups do not have subgroups of index 5.

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