My question is: If |G|=15, show that G must have an element of order 3.
We're currently learning about Lagrange's Theorem, so I know that I need to use it somehow, and I can't use Sylow's or Cauchy's Theorems yet. It makes sense that elements of order 3 and 5 (and 1 and 15) would be in G but I don't know how to use Lagrange to show that.
Thanks! Any help would be greatly appreciated!
Okay, so what I have so far is that
If G is cyclic, then it must contain some <g> where |g| = 3.
So, let G be non-cyclic. Then for all G, |g| = 1, 3, or 5. If |g|=1, it is the identity element. Assume |g|=5 for all g except the identity. Then G is cyclic which is a contradiction.
Or am I way off?