Assume that G is a group of order 63 that has two Sylow subgroups whose intersecion is non-trivial. Show that G has an element of order 21.
So by Sylow's theroem, I know that 63=3^2*7, that the sylow 7 subgroup is a normal subgroup of G with 6 order 7 elements. My sylow 3 subgroup is of order 9 and there can be one or 2 sylow three subgroups. Since I know that s7 (sylow 7 subgroup) is normal, I was thinking about trying to use this fact ot generate a cylic subgroup of order 21. Am I anywhere on the right path?
I'd check some non-trivial element in the intersection of two Sylow 3-subgroups.
So do I have to priove that their are 7 Sylow 3-subgroups, and from this I would get a subgroup of order 21, and since 7 is a prime and 3 is a prime they would both be in that subgroup? I am unsure how this implies that their would be an order 21 element though...
How would I prove that there is 7 3-sylows?
So since we have two sylow subgroups whose intersection is non-trival, ths means that we have 7 sylow 3 subgroups. This means we have 14 elements of order 3 in are group G. Since we have a unique sylow 7 subgroup, we have 6 elements of order 7 in G.
Let H = sylow 7 subgroup.
So for every a in G, aha^-1 is a subgroup of order 7, so we must have aHa^-1=H (since H is normal). So N(H) = G (N(H)= normalizer of H). SInce H has prime order, it is cyclic and abelian. So C(H) contains H (C(H)= centralizer of H in G). So 7 divides |C(H)| and |C(H)| divides 63. So C(H)= H or C(H)=G, or |C(H)|= 21.
Is this enough to get that xy=yx wth |x,y|=21?
Thanks again for your help.
If there's only one Sylow 3-sbgp. then it is normal, and since also the Sylow 7-sbgp. is normal and (9,7) = 1 , we get that G is the direct product of these two Sylow sbgps., which are each abelian, and thus G is abelian, and a finite abelian group has a subgroup of order any divisor of its order, and thus there's a sbgp. of G of order 21 which automatically is cyclic...
So the non-trivial case is when there's more than 1 Sylow 3-sbgp.