Sylow Subgroups: Show 10 is Not a Subgroup of Order 324

[beetle2]

Hi Guy's,
I know this is not for home work questions however I have had no luck in that section.

Have I done enough to show that 10 cannot be a sub-group of order 324


1. Homework Statement

Let G be a group of order 324. Show that G has subgroups of order 2,
3, 4, 9, 27 and 81, but no subgroups of order 10.


2. Homework Equations

Sylow showed that if a prime power divides the order of a finite group G, then G has a subgroup of order .

3. The Attempt at a Solution


I can see that G can have the subgroup 2 because [itex]2^n n=1 = 2[\latex]<br /> subgroup 3 because [itex]3^n n=1 = 3[\latex] divides 324<br /> subgroup 4 because [itex]2^n n=2 = 4[\latex] divides 324<br /> subgroup 9 because [itex]3^n n=2 = 9[\latex] divides 324<br /> subgroup 27 because [itex]3^n n=3 = 27[\latex] divides 324<br /> subgroup 81 because $3^n n=4 = 81[\latex] divides 324<br /> <br /> I know that 10 does not divide 324 in Z<br /> <br /> Is that enough to show that the sub group can't be order 10 ?$[/itex][/itex][/itex][/itex][/itex]

 

[Office_Shredder]

Yes, see Lagrange's theorem

 

[beetle2]

thanks a lot