Prove Finite Simple Group has no Subgroup of Index 5

[bham10246]

Question: Suppose that $G$ is a finite simple group of order greater than $5$ that contains no elements of order $2$. Prove that $G$ contains no subgroup of index $5$.

Attempt: Suppose G has a subgroup H of index 5, i.e., [G]=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 $|G|=(5^a) m$ where $5 \not | m$, then for $n_5 = \#(Sylow\: 5\: subgroups)$, $n_5 | m$ implies that $n_5$ is odd. If $n_5 =1$, this contradicts that G is simple. So $n_5 \geq 3$.

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

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

Thank you.

 

[bham10246]

I thought that given a finite group G and any subgroup H of G, we have [G]=|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]=|G|/|H|.

 

[matt grime]

The index of s subgroup is indeed just the number of cosets - there is no requirement for the subgroup to be normal at all.

 

[morphism]

A finite simple group is a finite group which has not non-trivial proper normal subgroup. So how it is possible for $$(G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />)=5$$ if that expression is only defined for groups $$H$$ so that $$H\triangleleft G$$ unless $$H=\{ e \}$$. 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]! (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.

 

[Kummer]

Question: Suppose that $G$ is a finite simple group of order greater than $5$ that contains no elements of order $2$. Prove that $G$ contains no subgroup of index $5$.

Attempt: Suppose G has a subgroup H of index 5, i.e., [G]=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 $|G|=(5^a) m$ where $5 \not | m$, then for $n_5 = \#(Sylow\: 5\: subgroups)$, $n_5 | m$ implies that $n_5$ is odd. If $n_5 =1$, this contradicts that G is simple. So $n_5 \geq 3$.

From Sylow's theorem, we also know that $n_5 \equiv 1 \mod 5$. So the last digit of $n_5$ 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 $$G=5^am$$ since $$n_5|G$$ it means $$n_5=5^bc$$ where $$c|m$$ and $$b\leq a$$. But $$5^bc$$ never ends in 1.

 

[morphism]

You got it. The last digit must be one. Now $$G=5^am$$ since $$n_5|G$$ it means $$n_5=5^bc$$ where $$c|m$$ and $$b\leq a$$. But $$5^bc$$ never ends in 1.

What if b=0 (which it must be, by Sylow)?

 

[Kummer]

What if b=0 (which it must be, by Sylow)?

Then we have,
$$c\equiv 1 (\bmod 5)$$ by Sylow's third theorem.
But $$c$$ 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.

 

[mathwonk]

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.

 

[morphism]

Yeah, that's basically the proof of the |G| divides [G]! 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.

 

[bham10246]

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!