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July 17th, 2012, 10:55 AM   #1
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The counting of p-sylow groups

Ok it says that for a group with order G, |G|=60, it has 6 5-sylow subgroups and 10 3-sylow subgroups, and 12,5 2-sylow subgroups. Ok i cant see why this is the case. Because , and say that it has a 5-sylow group, then by sylow's third theorem, then there are {1,6,11,......56} by congruence with 1mod(5) right? then the only divisible with the order of |G|=60 is 1 and 6, therefore 2 5-sylow subgroups. Ok maybe i am not doing it right, but isn't this is how you count it? i just dont see how its 6 5-sylow groups. Also how is there a two possibility of a 2-sylow for a order 60 group?

can anyone explain it to me? thank you
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July 17th, 2012, 10:35 PM   #2
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Re: The counting of p-sylow groups

a 5-sylow subgroup of a group G of order 60 has order 5. of course, these all intersect in the identity, so if k is the number of 5-sylow subgroups of G, these account for 4k+1 elements of G.

by the third sylow theorem, k = 1 (mod 5), and k divides 60, so either k = 1, or k = 6. note that if k = 6, the 5-sylow subgroups only account for 25 elements of G, leaving 35 elements of other orders.

let's show that both scenarios actually occur. for a single 5-sylow subgroup, consider (Z60,+), the integers mod 60 under addition modulo 60. the sole subgroup of order 5 is:

{0,12,24,36,48}

for 6 5-sylow subgroups, consider A5. there are 24 (= 4!) 5-cycles in S5, and all of these are in A5. explicitly we have:

<(1 2 3 4 5)>
<(1 2 3 5 4)>
<(1 2 4 3 5)>
<(1 2 4 5 3)>
<(1 2 5 3 4)>
<(1 2 5 4 3)>, which are 6 distinct subgroups of order 5 in A5.

of course a 2-sylow subgroup is possible. such a subgroup has (for a group of order 60) order 4. for a group of order 60, we can have 1,3 5, or 15 of them (these being the odd divisors of 60/4 = 15). again, using our two examples, we see Z60 has just one subgroup of order 4:

{0,15,30,45}

while A4 has the following:

{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}
{e, (1 2)(3 5), (1 3)(2 5), (1 5)(2 3)}
{e, (1 2)(4 5), (1 4)(2 5), (1 5)(2 4)}
{e, (1 3)(4 5), (1 4)(3 5), (1 5)(3 4)}
{e, (2 3)(4 5), (2 4)(3 5), (2 5)(3 4)} which accounts for all possible subgroups of order 4 in A5 (two double 2-cycles which don't move the same 4 elements result in a 3-cycle when multiplied), so we have 5 2-sylow subgroups.

you might want to hazard a guess as to how many 2-sylow subgroups D30 (the dihedral group of order 60) has.
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