How many abelian groups of order $p^{5}$ are there? I want to find how many abelian groups of order $p^{5}$ there are, up to isomorphism, where $p$ is a prime number.  Are there any theorems that can help with this?

Edit: Adding my answer of your question in comments:
To use the theorem you must factor suppose $\vert G \vert=n$ and factor $n$ into prime factors. Then you must get divisors $d_1,d_2,\dots,d_m$ such that $d_i\mid d_{i+1}$, and $d_1\cdot\dots\cdot d_{m}=n$, and your group can be
$$G\cong \mathbb{Z}_{d_1}\oplus \mathbb{Z}_{d_2}\oplus \dots \oplus \mathbb{Z}_{d_m} $$
Original answer
Yes, the theorem that you're looking for is the fundamental theorem of finitely generated abelian groups. From this you have that $p^5$ can factor into 
$$p^5=p^5$$ 
$$p^5=p\cdot p^4$$ 
$$p^5=p^2\cdot p^3$$ 
$$p^5=p\cdot p\cdot p^3$$ 
$$p^5=p\cdot p^2\cdot p^2$$ 
$$p^5=p\cdot p\cdot p\cdot p^2$$ 
$$p^5=p\cdot p\cdot p\cdot p\cdot p$$ 
So you have all the direct sums of that groups.
$$G\cong\mathbb{Z}_{p^5}$$ 
$$G\cong\mathbb{Z}_{p}\oplus \mathbb{Z}_{p^4}$$ 
$$G\cong\mathbb{Z}_{p^2}\oplus \mathbb{Z}_{p^3}$$ 
$$G\cong\mathbb{Z}_{p}\oplus \mathbb{Z}_{p}\oplus \mathbb{Z}_{p^3}$$ 
$$G\cong\mathbb{Z}_{p}\oplus \mathbb{Z}_{p^2}\oplus \mathbb{Z}_{p^2}$$ 
$$G\cong\mathbb{Z}_{p}\oplus \mathbb{Z}_p\oplus \mathbb{Z}_{p}\oplus \mathbb{Z}_{p^2}$$ 
$$G\cong\mathbb{Z}_{p}\oplus \mathbb{Z}_p\oplus \mathbb{Z}_p\oplus \mathbb{Z}_{p}\oplus \mathbb{Z}_{p}$$ 
