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Let $p$ be a prime number. Find the number of abelian groups of order $p^n$, up to isomorphism when $n=2,3$, and $5$.

I know the answer when $n=2$ and $3$. And my professor said that there are $7$ abelian groups when $n=5$ but I do not understand where he got that number.

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Use the structure theorem of finitely generated abelian groups - this becomes a partition number problem. – mixedmath Apr 28 '14 at 8:48

Use the fundamental theorem of finitely generated abelian groups. This gives us $7$ groups:

$$\begin{array}{c} \mathbb Z_{p^5} \\ \mathbb Z_{p^4} \times \mathbb Z_{p} \\ \mathbb Z_{p^3} \times \mathbb Z_{p^2} \\ \mathbb Z_{p^3} \times \mathbb Z_{p} \times \mathbb Z_p \\ \mathbb Z_{p^2} \times \mathbb Z_{p^2} \times \mathbb Z_p \\ \mathbb Z_{p^2} \times \mathbb Z_p \times \mathbb Z_{p} \times \mathbb Z_p \\ \mathbb Z_p \times \mathbb Z_{p} \times \mathbb Z_p \times \mathbb Z_{p} \times \mathbb Z_p \end{array} $$

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