# In a finite abelian group of order n, must there be an element of order n?

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.

-
Are you really asking if all abelian groups are cyclic? – lhf Nov 2 '12 at 21:21
"really" here is read "really?!" – rschwieb Nov 2 '12 at 21:24
I wasted at least an hour today for not seeing something as simple as this implication. But this brings me backwards in another demonstration I am attempting, which I will now post. – Temitope.A Nov 2 '12 at 21:29

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.

-
"If a group has order $n$ and $p$ is a prime dividing $n$, then there is an element of the group of order $p$". This is known as Cauchy's Theorem. – BobaFret Nov 2 '12 at 21:38

No: consider the Klein $4$-group.

-

Consider any finite noncyclic abelian group. Any one at all. Take your pick!

-
@tomasz No, like one that satisfies the hypotheses of the question! (Thanks for making me type it in.) – rschwieb Nov 2 '12 at 21:27