In Fraleigh, there is a true/false problem in the chapter on "Direct Products" that states the following three assertions:
- Every abelian group of order divisible by 5 contains a cyclic subgroup of order 5.
- Every abelian group of order divisible by 4 contains a cyclic subgroup of order 4.
- Every abelian group of order divisible by 6 contains a cyclic subgroup of order 6.
I know the second statement is false because, for example, the Klein-4 group does not contain a subgroup of order 4. The solutions key says that 1 and 3 are true, but I'm not sure why (there is no explanation). Is this somehow a consequence of the Fundamental Theorem of Finitely Generated Abelian Groups?
Best Answer
Every Abelian group can be decomposed into into the direct sum of cyclic groups of prime power order.
A group of order divisible by $5$ has some cyclic subgroup of order $5^n$ take the generator of that subgroup call it $a$ and $a^n$ is the generator of a group of order $5$
A group of order divisible by $6$ has some subgroup of order $2$ and a subgroup of order $3$.
Take the generator of both and multiply them together. Since $gcd(2,3) = 1$ this will generate a group of order $6.$