I have been attempting to create the subgroup lattice for $\Bbb{Z}_4 \times \Bbb{Z}_8$. I have, so far, this: http://www.scribd.com/doc/223680804/Subgroup-Lattice-of-Z-4-x-Z-8
While I have calculated every single subgroup by hand in order to create this lattice, my intuition tells me that something is off because the top of the lattice doesn't seem symmetrical to the bottom nor are is the left half a mirror of the right half. While this isn't necessarily a requirement, most subgroup lattices of $\Bbb{Z}_m \times \Bbb{Z}_n$ I've encountered look more symmetrical than this.
Can anyone verify that this is correct or show me where my error is located?
EDIT: I have edited the subgroup lattice to include all isomorphisms. If anyone wants me to provide any further information pertaining to my lattice, just ask and I shall post it.
You seem to have missed the difference between $\Bbb Z_k$ and a general abelian group of order $k$. For instance, $\Bbb Z_8\times\Bbb Z_4$ does indeed have seven subgroups of order $4$, but one of them (the middle dot on your $\Bbb Z_4$-row) is actually $\Bbb Z_2\times\Bbb Z_2$, not $\Bbb Z_4$. Similarly, only four of your $\Bbb Z_8$s are really $\Bbb Z_8$s - the other three are $\Bbb Z_4\times\Bbb Z_2$. There are two (not three) subgroups isomorphic to $\Bbb Z_8\times\Bbb Z_2$, and none isomorphic to $\Bbb Z_{16}$ (since that group has an element of order $16$, but $\Bbb Z_8\times\Bbb Z_4$ doesn't).
Once you rethink the subgroups accordingly, you should be able to get a more accurate subgroup lattice. It definitely won't be symmetric top-to-bottom. Whether it's symmetric left-to-right might be an artifact of how you draw it.