Hi Guy's,(adsbygoogle = window.adsbygoogle || []).push({});

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. The problem statement, all variables and given/known data

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. Relevant 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]

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 ?

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# Sylow Subgroups

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