I need a little help with understanding homology groups.
In particular, consider a simplicial complex with three 1-cycles and one 1-boundary (so we have two 1-holes?). Then the first homology group is the second order cyclic group, i.e. $H_1\tilde{=}\frac{\mathbb{Z}^3}{\mathbb{Z}}\tilde{=}\mathbb{Z}^2$. Is this correct?
If so, the problem is I'm getting $\frac{\mathbb{Z}^3}{\mathbb{Z}}=\mathbb{Z}$ here.
You should indeed be getting $\mathbb Z^2$ here. One way to see this is that we can shrink the "filled" triangle down to a point, leaving you with two circles meeting at a single point, whose homology is $\mathbb Z^2$.
The cycles are the free abelian group generated by $[ab]+[bc]+[ca]$, $[bc]+[cd]+[db]$ and $[db]+[be]+[ed]$. The boundaries are the subgroup generated by $[bc]+[cd]+[db]$, so you get $\mathbb Z^2$ as expected.
I don't know exactly where you went wrong in this calculation because you didn't provide much of your working, but if you're more explicit I can try to answer your question more thoroughly.