# Why are certain combinations of factors left out in the list of abelian groups of order 24 up to isomorphism?

I'm trying to find all the abelian groups of order $24$ up to isomorphism. To that end, I am using the following result: Every finitely generated abelian group is isomorphic to a finite direct sum $\mathbb{Z}_{d_{1}}\oplus \mathbb{Z}_{d_{2}} \oplus \cdots \oplus \mathbb{Z}_{d_{k}} \oplus \mathbb{Z}^{m}$ where $d_{i}>1$ and $d_{i}\vert d_{i+1}$ for all $i$.

Then, as an immediate corollary to this, we have that if $d=p_{1}^{r_{1}}p_{2}^{r_{2}}\cdots p_{k}^{r_{k}}$ is a primary decomposition of a number $d$ then $\mathbb{Z}_{d}\simeq \mathbb{Z}_{p_{1}^{r_{1}}}\oplus \mathbb{Z}_{p_{2}^{r_{2}}} \oplus \cdots \oplus \mathbb{Z}_{p_{k}^{r_{k}}}$ where $\mathbb{Z}_{p^{r}}$ is a primary cyclic group.

Now, I've heard that the only abelian groups of order $24$ up to isomorphism are

1. $\mathbb{Z}_{2}\oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3}$
2. $\mathbb{Z}_{2}\oplus \mathbb{Z}_{4} \oplus \mathbb{Z}_{3}$
3. $\mathbb{Z}_{8}\oplus \mathbb{Z}_{3}$

So, my question is why is this list exhaustive? In other words, why are things like $\mathbb{Z}_{4}\oplus \mathbb{Z}_{6}$ and $\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{6}$ not included? As far as I know, they're abelian, they're order $24$. Are they isomorphic to one or more of the listed groups? Is it because we must have $d_{i}\vert d_{i+1}$ for all $i$? (Although then how does this explain $\mathbb{Z}_{2} \oplus \mathbb{Z}_{4} \oplus \mathbb{Z}_{3}$, since $4$ does not divide $3$?)

I've done a bit of searching online and found something called the Fundamental Theorem of Abelian Groups, although in my course, we have never learned it. Probably the closest thing to it that we have learned is the result I mentioned above about finitely generated abelian groups (and of course, a finite abelian group is finitely generated). So, could somebody please explain to me in a detailed way why this list of abelian groups of order $24$ is exhaustive? Thank you in advance.

Because $2$ and $3$ are coprime, $\mathbb Z_6\simeq\mathbb Z_2\oplus\mathbb Z_3$. So the group $\mathbb Z_4\oplus\mathbb Z_6$ is number 2 in your list, and the group $\mathbb Z_2\oplus\mathbb Z_2\oplus\mathbb Z_6$ is number 1 in your list.
• also, do the Sylow $p$ theorems have anything to do with this? – ALannister Mar 14 '17 at 18:55