# An example of a group that that has an order of M that is abelian?

Theory: If G is a ﬁnite abelian group, p is prime and p divides the order of G then G has an element of order p.

Can anyone think of a counter example for a number n that is not prime, divides the order of some group, and the result proves its false.

I have looked at cyclic groups, the group of nth roots of unity and every finite abelian group I know, and no matter what, if some number divides the order of a group it has some element in it that has an order of that number. It doesn't really matter if it's prime or not. Any examples?

## 3 Answers

The Klein group is abelian and order four, but no element has order four. More generally, a group that is the direct product of two cyclic groups that have orders that are not coprime will not have an element of the order of the group.

what would be the order of any element in $\Bbb Z_8^*=\{1,3,5,7\}$, so there is no element of order $4$

Consider the direct product of, say, four cyclic groups of order 3, which has order 81 and is abelian. Does it have elements of order $9$, or 27?