If $p$ is prime, determine the number of abelian groups of order $p^n$ for each $1\leq n\leq8$
(I assume that "up to isomorphism" should be included somewhere in the question for the sake of precision...)
Could someone please review/confirm my work?
n = 1: $\mathbb{Z}_p $
n = 2: $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_p\times \mathbb{Z}_p$
n = 3: $\mathbb{Z}_{p^3}$, $\mathbb{Z}_{p^2}\times \mathbb{Z}_p$, and $\mathbb{Z}_p\times \mathbb{Z}_p \times\mathbb{Z}_p$
n = 4: $\mathbb{Z}_{p^4}$, $\mathbb{Z}_{p^3} \times \mathbb{Z}_p$, $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p^2}$, $\mathbb{Z}_{p^2}\times \mathbb{Z}_p \times \mathbb{Z}_p$, and $\mathbb{Z}_p\times \mathbb{Z}_p \times\mathbb{Z}_p \times \mathbb{Z}_p$
et cetera
I am simply considering all the options for when the largest exponent of $p$ is $n$, then $n-1$, and so on. How does this look? Thanks!
(Apparently I don't know how to "end a quote"...)