Krdytkyu

In reading about group cohomological, I came across the following in a statement of a lemma. "Let $p$ be a prime, and let $A$ be an abelian group of exponent dividing $p$..."

Is this just a roundabout way of saying that $A$ is an $mathbb{F}_p$-vector space? At least in the case where $A$ is finite or finitely generated, this seems obvious from the classification of finitely generated abelian groups. According to Exponent of a Group, I am correct, but perhaps I am missing an assumption there.

Are there abelian groups of exponent $p$ that are not $mathbb{F}_p$-vector spaces?

Yes, this is the same thing as an $mathbb{F}_p$-vector space. There is no need to think about any classification theorem; this is instead just immediate from the definitions. Since $mathbb{F}_p$ is the quotient ring $mathbb{Z}/(p)$, an $mathbb{F}_p$-module is the same thing as a $mathbb{Z}$-module in which every element is annihilated by $p$. But a $mathbb{Z}$-module is the same thing as an abelian group, and every element of an abelian group is annihilated by $p$ iff the exponent of the group divides $p$.