If G 9 can have only elements of order 139SolutionSylows fi

If |G| = 9 can have only elements of order 1,3,9?

Solution

Sylow\'s first theorem states that if pⁿ is the maximal power of p dividing the order of G, then G has a subgroup of order pⁿ, then, using the fact that a p-group is solvable, one can show that G has subgroups of order any power p^r of p dividing n).As per Cauchy’s theorem, if G is a finite group and p is a prime number dividing the order of G, then G contains an element of order p. Let G be a group and g G. We say g has finite order n if n is the smallest positive integer for which gⁿ = e ( the identity element of G).

Since 3² is the maximal power of the prime number 3 dividing 9 ( the order of G), therefore, in view of the above, G can only have elements of the order 3⁰ , 3¹ and 3³ i.e. 1, 3 and 9.