Finding sylow subgroups

[shakeydakey]

Homework Statement

Write down all Sylow subgroups of A5 (alternating group). I have them for order 3 and order 5 subgroups, but am not sure about order 4 subgroups.

Homework Equations

Sylow Theorems etc

The Attempt at a Solution

I know the # of 4-subgroups is congruent to 1mod2 (though that's not helpful), and a factor of 60/4=15, from Sylow Thm 3. So it's 1,3,5, or 15. I've found 5 of the form {(ab)(cd), (ac)(bd),(ad)(bc),identity}. If there are more I can't find them, and if that's it how do I show that?

 

[mighty2000]

Dear fellow scientist,

Thank you for your post. I am happy to assist you in finding the Sylow subgroups of A5. First, let's review the Sylow theorems that will be useful in this problem.

1. Sylow's First Theorem: For any prime number p, if p^k is the highest power of p that divides the order of a finite group G, then G contains a subgroup of order p^k.
2. Sylow's Second Theorem: If p^k is the highest power of p that divides the order of a finite group G, then any two subgroups of G of order p^k are conjugate.
3. Sylow's Third Theorem: If p^k is the highest power of p that divides the order of a finite group G, then the number of subgroups of G of order p^k is congruent to 1 mod p.

Now, let's apply these theorems to A5. The order of A5 is 60, which can be factored as 2^2 * 3 * 5. Using Sylow's Third Theorem, we can determine the number of Sylow 2-subgroups, 3-subgroups, and 5-subgroups.

For 2-subgroups: The number of 2-subgroups is congruent to 1 mod 2 and a factor of 15. Therefore, the only possibility is 1 2-subgroup of order 4. We can easily find this subgroup as {(ab)(cd), (ac)(bd), (ad)(bc), identity}.

For 3-subgroups: The number of 3-subgroups is congruent to 1 mod 3 and a factor of 20. Therefore, there are either 1 or 5 3-subgroups. We can find one 3-subgroup as {(abc), (acb), identity}. To find the other 3-subgroups, we can use Sylow's Second Theorem to show that they are all conjugate.

For 5-subgroups: The number of 5-subgroups is congruent to 1 mod 5 and a factor of 12. Therefore, there are either 1 or 3 5-subgroups. We can find one 5-subgroup as {(abcde), identity}. Again, we can use Sylow's Second The