I am trying to find an example of a nonabelain group of order 100 in which every element has order at most 10.
I have been trying to use dihedral groups (D_n = {e, f, f^2, … , f^{n-1}, g, fg, f^{2}g, …, f^{n-1}g} ) and have considered the dihedral group D_50. This would have 100 elements so it would have order 100 although I am not entirely sure how to show that each element has order of at most 10.
If anyone has any suggestions on how to prove this or any other suggestions of examples that would be extremely helpful!
Best Answer
It seems $D_5 \times D_5$ should work. $100$ elements, non-abelian. Each element in the base groups can have order $1,2,5,10$ so the order of an element in the product group, which is the least common multiple of the orders in the base groups, is at most $10$.