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

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beetle2
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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 2^n n=1 = 2[\latex]<br /> subgroup 3 because 3^n n=1 = 3[\latex] divides 324&lt;br /&gt; subgroup 4 because 2^n n=2 = 4[\latex] divides 324&amp;lt;br /&amp;gt; subgroup 9 because 3^n n=2 = 9[\latex] divides 324&amp;amp;lt;br /&amp;amp;gt; subgroup 27 because 3^n n=3 = 27[\latex] divides 324&amp;amp;amp;lt;br /&amp;amp;amp;gt; subgroup 81 because 3^n n=4 = 81[\latex] divides 324&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; &amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; I know that 10 does not divide 324 in Z&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; &amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; Is that enough to show that the sub group can&amp;amp;amp;amp;amp;#039;t be order 10 ?
 
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Yes, see Lagrange's theorem
 
thanks alot
 

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