Solving Group Theory Problems: Sylow, Abelian, and Order 36

[basukinjal]

1. Let G be a fintie group whose order is divisible by a prime p. Assume that (ab)^p = a^p.b^p for all a,b in G. Show that the p-Sylow subgruop of G is normal in G.

2. Find the number of Abelian groups of order 432.

3. Let G be a group of order 36 with a subgroup H of order 9. Show that H is normal in G.

 

[quantumdude]

basukinjal, please stop deleting the posting template that appears when you start a new thread in the homework help forum. You must use it, and you must fill in the third section, which is entitled "Attempt at a solution." If you don't try the problem, we won't help you. Those are the rules that you agreed to upon registration.

 

[basukinjal]

ok. Sorry for that, here are my attempts.

1. Since (ab)^p = a^p.b^p, i tried to construct a homomorphism phi, such that phi(x) = x^p. Then the kernel for this would not be just e since p | o(G) thus, this is not a isomorphism.. and i got stuck there after.

2. 432= 2^4*3^3. So i tried to construct the 2-sylow subgroup and the 3-sylow subgroup, but in that case no. of 3- sylow subgroups = 1 + 3k which must divide 16. thus there can be 1,4 or 16 sylow 3-subgroups, similarly, for 2-sylow subgroups there must be 1,3,9 or 27 sylow 2-subgruops... what then??

3. If we can prove that H is the only sylow 3-subgroup we are done. No. of sylow 3-subgroups are 1+3k which must divide 4. if 1, my problem is solved. otherwise there can be 4 sylow 3- subgroups. we have to show that this cannot be the case. So N(P) where P is any sylow 3-subgroup has index 4. i also noted the fact that o(G)=36 does not divide i(N(P))!, thus there must be a non trivial normal subgroup in N. But then i cannot prove that H has to be normal.

Please help

 

[Office_Shredder]

2 doesn't really look like a Sylow subgroup question; have you done anything regarding the decomposition of Abelian groups into a direct product of cyclic groups?