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Let $G$ be a finite simple group with $|G|\geq 6$. For each of the following statements, prove or provide a counterexample.

  1. $G$ cannot have an index-$2$ subgroup.
  2. $G$ cannot have an index-$3$ subgroup.
  3. $G$ cannot have a proper subgroup of index less than $6$.
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One can show that any index $2$ subgroup must be normal. Hint: What are the left cosets? What are the right cosets? So by that result, you answer 1), and see why looking at $S_4$ or any $S_n$ won't help in this problem: $A_n$ is always a normal subgroup, so $S_n$ is not simple. –  Ragib Zaman Jul 3 '12 at 2:44
    
@RagibZaman Thanks! I forgot I am working only with simple groups. Also with regards to your comment "So by that result, you answer 1)" I don't see how. Can you please elaborate? Thanks. –  Lyapunov Jul 3 '12 at 2:52
    
If G had an index 2 subgroup, that subgroup would be normal, contradicting that G is a simple group. –  Ragib Zaman Jul 3 '12 at 2:53
    
@RagibZaman Yes! I should go to bed! –  Lyapunov Jul 3 '12 at 2:54
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@POTUS: You have two hypothesis: $G$ is simple, and $|G|\geq 6$. The way Ragib wrote his statement, "If $G$ had a subgroup of index $2$ then that subgroup would be normal, contradicting that $G$ is simple" is not correct. Simple groups do have normal subgroups: they just can't be proper. So the correct full argument is tat if $G$ has a subgroup of index $2$, then the subgroup is normal. Since $G$ is simple, that subgroup must be trivial. Therefore, $G$ must have two elements, contradicting that $|G|\geq 6$ (which was the "latter hypothesis"). –  Arturo Magidin Jul 3 '12 at 3:17
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1 Answer 1

up vote 8 down vote accepted

Let $G$ be a group. If $H$ is a subgroup of index $n$, then $H$ contains a normal subgroup of $G$ of index at most $n!$.

This is a common problem; consider the action of $G$ on the left cosets of $H$ by left multiplication, $g\cdot xH \longmapsto gxH$. This induces a homomorphism $G\to S_n$, and the kernel is contained in $H$. Since $G/K$ is isomorphic to a subgroup of $S_n$, the index of $K$ is at most $n!$.

If $G$ is assumed to be simple, then $K$ must be trivial.

This will tell you the only possible counterexamples for index 2 or 3. Try them and see if they meet all your hypothesis.


Added. Assume that $G$ is a simple group, and $H$ is a subgroup of index $2$ in $G$. Then $H$ contains a subgroup $K$ that is normal in $G$, and has index at most $2!$. Since $K\subseteq H$ and $[G:H]=2$, then $[G:K] = [G:H][H:K] = 2[H:K]\geq 2$, it follows that $[G:K]=2$, so $H=K$. Since $G$ is simple and $H$ is normal, it follows that $H=\{e\}$. But since $[G:H]=2$, it follows that $|G|=2$. That is, $G$ is the cyclic group of order $2$. In summary: the only simple group that has a subgroup of index $2$ is the cyclic group of order $2$.

Thus, if $G$ is a simple group and $|G|\geq 6$, then $G$ cannot have a subgroup of index $2$. Thus, 1 is true.

Assume $G$ is a simple group and $H$ is a subgroup of index $3$ in $G$. By the result above, $H$ contains a normal subgroup $K$ of $G$ that has index at most $3!$ in $G$. Thus, $K=\{e\}$ by the same argument as above, so $|G|= [G:K] \leq 6$. The only simple groups of order at most $6$ are $C_2$, $C_3$, and $C_5$; the only one that has a subgroup of index $3$ is $C_3$. So $G$ will be cyclic of order $3$.

Thus, if $G$ is a simple group of order at least $6$, then $G$ cannot have a subgroup of index $3$. Thus, 2 is true.

For subgroups of index $4$, we get as above that $G$ contains a normal subgroup of index at most $4! = 24$; this must be trivial, so $|G|\leq 24$. The only simple groups of order at most $24$ are $C_p$ with $p$ a prime, $p\leq 24$, and none of them has a subgroup of index $4$. So no simple group has a subgroup of index $4$.

Subgroups of index $5$ yield a normal subgroup of index at most $5!= 120$. Now we have some possibilities, since there are nonabelian groups of order less than $120$. For instance, we have $A_5$, of order $60$. Does $A_5$ have a subgroup of index $5$? That would be a subgroup of order $12$. And in fact, $A_5$ does contain subgroups of order $12$, for example, the stabilizer of a point is isomorphic to $A_4$, of order $12$. Thus, $A_5$ is a simple group of order at least $6$ that has a subgroup of index $5$. So 3 is false (though "barely", as it were, since a simple group as given does not have subgroups of indices 2, 3, or 4).


$S_4$ does you no good, since $S_4$ has lots of normal subgroups: $A_4$, and also a normal subgroup of order $4$.

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I don't think I need counterexamples for 1). 1) is true by the above discussion, so I need prove it, right? –  Lyapunov Jul 3 '12 at 3:17
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@POTUS: The point is that the only simple group that has a subgroup of index $2$ is the group of order $2$; but that group does not satisfy your hypothesis that $G$ has order at least $6$. For part 2, you conclude that $|G|=6$ (since the normal subgroup guaranteed by the result quoted above has index at most $6$, but must be trivial). But there are no simple groups of order $6$, so this cannot occur. For index $4$, $5$, or $6$, you'll want to think about it a bit more. –  Arturo Magidin Jul 3 '12 at 3:21
    
I am still confused, can you clarify which of the statements are true and which are not? Right now I feel like you are arguing that it is and it is not! –  Lyapunov Jul 3 '12 at 3:37
    
@POTUS: Hrmph. I see why you are confused; I keep saying "false" when I meant "true". –  Arturo Magidin Jul 3 '12 at 5:25
    
Okay! Phew! Thank you for being patient with me though! –  Lyapunov Jul 3 '12 at 5:43
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