I've got a finite, abelian group and I know the order of all of the elements.
How far does that get me towards knowing what the group is up to isomorphism?
The group has eight elements. Excluding the identity, three elements have order two, while four elements have order four.
I also have the group tables, and I know that I could shuffle the rows and columns looking for some patterns to emerge, but I'd rather have a theoretical approach.
On
The fundamental theorem of finitely generated abelian groups is something of a sledgehammer for this problem, but anyways: any abelian group of order $8$ is isomorphic to one of $\mathbb{Z}/8\mathbb{Z}, \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, or $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Your group has no element of order $8$, which rules out the first case, and every element of $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ has order $2$, so your group must be $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.
That tells you everything: take an element $a$ of order 4, look at the subgroup it generates; find a non-identity element $b$ in the complement of this group. If $b$ has order 2, keep it; if not, square it and it'll have order 2. At this point, your group is generated by $a$ and $b$ with $a^4 = b^2 = e$ and $ab = ba$. And that's the complete structure.
In your particular group, $C$ has order $4$, with the subgroup generated by it being $$ Id, C, B, D $$ and so $A$, in the complement of this subgroup, serves as the other generator. Your group is (or is isomorphic to)
$$ \mathbb Z/ 4\mathbb Z \oplus \mathbb Z/ 2\mathbb Z $$ with the first factor generated by $C$ and the second by $A$. Let me make that more explicit:
Consider the map \begin{align} Id &\mapsto (0,0) \\ C &\mapsto (1,0) \\ B &\mapsto (2,0) \\ D &\mapsto (3,0) \\ A &\mapsto (0,1) \\ AC = E &\mapsto (1, 1)\\ AB = G & \mapsto (2, 1)\\ AD = F &\mapsto (3, 1) \end{align} That's an explicit isomorphism of groups.