Showing Subgroups of Order 2, 3, 4, 9, 27 and 81 in Group of Order 324

[beetle2]

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.

Homework Equations

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

The Attempt at a Solution

I can see that G can have the subgroup 2 because [itex]2^n n=1 = 2[\latex]
subgroup 3 because [itex]3^n n=1 = 3[\latex] divides 324
subgroup 4 because [itex]2^n n=2 = 4[\latex] divides 324
subgroup 9 because [itex]3^n n=2 = 9[\latex] divides 324
subgroup 27 because [itex]3^n n=3 = 27[\latex] divides 324
subgroup 81 because [itex]3^n n=4 = 81[\latex] divides 324

I know that 10 does not divide 324 in Z

Is that enough to show that the sub group can't be order 10 ?

 

[mighty2000]

Yes, your reasoning is correct. Since 10 does not divide 324, there cannot be a subgroup of order 10 in G. This is because, as you mentioned, Sylow's theorem states that if a prime power p^k divides the order of G, then G has a subgroup of order p^k. Since 10 is not a prime power that divides 324, there cannot be a subgroup of order 10 in G.