1. ## Simple group question

I need some direction with some of the following:

Let $\displaystyle G$ be a simple group $\displaystyle |G|=60$

1. show that $\displaystyle G$ does not have sub-groups of index 3 or 4.
2. show that if $\displaystyle H_{1},H_{2}$ are 2 different 2-sylow subgroups of $\displaystyle G$ such that $\displaystyle H_{1}\cap H_{2}\neq \{e\}$ then $\displaystyle N(H_{1}\cap H_{2})$ is a subgroup of index 5.
3. show that if for every 2-sylow subgroups the intersection is trivial, $\displaystyle H_{1}\cap H_{2} = {e}$. then there are exactly five 2-sylow subgroups.
4. show that $\displaystyle G$ is isomorphic to $\displaystyle A_{5}$.

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.

2. To get you started, note that $\displaystyle G$ acts on its Sylow 2-subgroups. If there were 3 of them, then this would induce a homomorphism $\displaystyle \phi:G\to S_3$. On one hand, $\displaystyle \ker \phi$ should not be trivial, because $\displaystyle |S_3|=6$, which is too small. On the other, $\displaystyle \ker \phi$ should not be $\displaystyle G$, because the action is nontrivial. This means that $\displaystyle \ker \phi$ is a proper nontrivial subgroup of $\displaystyle G$. Since the kernel is always normal, this contradicts the simplicity of $\displaystyle G$.

Try using a similar trick for part four.

3. Originally Posted by roninpro
To get you started, note that $\displaystyle G$ acts on its Sylow 2-subgroups. If there were 3 of them, then this would induce a homomorphism $\displaystyle \phi:G\to S_3$. On one hand, $\displaystyle \ker \phi$ should not be trivial, because $\displaystyle |S_3|=6$, which is too small. On the other, $\displaystyle \ker \phi$ should not be $\displaystyle G$, because the action is nontrivial. This means that $\displaystyle \ker \phi$ is a proper nontrivial subgroup of $\displaystyle G$. Since the kernel is always normal, this contradicts the simplicity of $\displaystyle G$.

Try using a similar trick for part four.

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.

For OP,

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 $\displaystyle |G:N_G(P)|=$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, $\displaystyle |G:N_G(P)|=5$. 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 $\displaystyle G \leq A_5$ and conclude that $\displaystyle G = A_5$.

4. Originally Posted by TheArtofSymmetry
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.
I should have been more explicit, but $\displaystyle G$ acts on its Sylow p-subgroups by conjugation (which follows by #3 in your list).

5. .

6. Thank you both