Simple group question
I need some direction with some of the following:
Let be a simple group
1. show that does not have sub-groups of index 3 or 4.
2. show that if are 2 different 2-sylow subgroups of such that then is a subgroup of index 5.
3. show that if for every 2-sylow subgroups the intersection is trivial, . then there are exactly five 2-sylow subgroups.
4. show that is isomorphic to .
I was able to show 1,2.
with 3, I am having trouble. using sylow's 3rd theorem I can show that the number of 2-sylow subgroups can be 1,3,5 or 15. I can eliminate the possibility of 1 (contradiction to simple group) and the possibility of 15 (too many elements considering the other p-sylow subgroups). but I can't eliminate the possibility of 3.
I need direction as weel on how to start 4.
To get you started, note that acts on its Sylow 2-subgroups. If there were 3 of them, then this would induce a homomorphism . On one hand, should not be trivial, because , which is too small. On the other, should not be , because the action is nontrivial. This means that is a proper nontrivial subgroup of . Since the kernel is always normal, this contradicts the simplicity of .
Try using a similar trick for part four.
Originally Posted by roninpro
Do you mean G acts on the set of left cosets of a Sylow 2-subgroup of G? If G acts simply on its Sylow-2 subgroups, I don't think the action is well-defined for this problem.
Some useful theorems for this problem are
1) If H is a subgroup of index n in a group G and G is simple, then G is isomorphic to a subgroup of S_n. (Think about G acts on the set of all left cosets of H in G)
2) The number of conjugates of H in G is |G:N_G(H)|
3) Any two Sylow p-subgroup of G are conjugate.
By (1), G does not have index 3 or 4 subgroup, since 3! or 4! is smaller than 60.
The possible numbers of Sylow 2-subgroups are 3, 5, or 15.
Let P be a Sylow 2-subgroup. Then 3, 5, or 15 by (2) and (3). 3 is not a possible number by (1). 15 is not also a possible number either by counting arguments. (Verify this).
Thus, the number of Sylow 2-subgroups are 5. Then, . By (1), we see that G is isomorphic to the subgroup of S_5 and is simple of order 60 by assumption. Argue first that and conclude that .
I should have been more explicit, but acts on its Sylow p-subgroups by conjugation (which follows by #3 in your list).
Originally Posted by TheArtofSymmetry