In a finite abelian group of order n, must there be an element of order n?
This question is bugging me, I've been thinking about it all afternoon long. Can somebody hint me? I guess it is true, but can find a clear path to the demonstration.
In a finite abelian group of order n, must there be an element of order n?
This question is bugging me, I've been thinking about it all afternoon long. Can somebody hint me? I guess it is true, but can find a clear path to the demonstration.
Consider any finite noncyclic abelian group. Any one at all. Take your pick!
No: consider the Klein $4$-group.
This is not true, as seen by ${\bf Z}_2^2$, for instance (or any finite non-cyclic abelian group, actually). If you can't prove a hypothesis, you should try to find a counterexample, whatever your intuition on it might be. If you find one, you will know why you couldn't prove it, and if you don't, the problems you find with finding a counterexample can be a good hint as to how to prove your hypothesis.
The simplest form of converse of Lagrange's theorem that I know and which actually works is that if a group has order $n$ and $p$ is a prime dividing $n$, then there is an element of the group of order $p$ (whether the group is abelian or not); this is known as Cauchy's theorem.