Lattice of Groups Let M and N be normal subgroups in G such that G=MN. Prove that G/(M intersection N) is isomorphic to G/M * G/N. Draw the lattice.
I solved the problem using first isomorphism theorem. But how can I draw the lattice. Can anyone help?
 A: Drawing the lattice of $G/N\times G/M$ is not such a trivial task (some nontrivial subgroups might arise in the direct product), but drawing the lattice of $G/(M\cap N)$ is rather easy: it is the lattice obtained by looking "up" from $M\cap N$. As an example, consider $Q_8$, of order $8$. If you take the subgroups generated by $i$ and $j$ respectively, the quotients are $C_2$, and the intersection is isomorphic to $C_2$. In both cases, of course, the product and the quotient are isomorphic to $V_4=C_2\times C_2$. Note that the lattice of $C_2$ is a chain of two elements, but the lattice of $C_2\times C_2$ has a Hasse diagram of five elements: $0$, three uncomparable subgroups of order $2$, and $C_2\times C_2$ as its top element. The lattice might be obtained easily from that of $Q_8$ by looking at the lattice above $C_2$. The lattice of $Q_8$ is just like that of $C_2\times C_2$; but has an extra element: to draw it, simply add a new (smaller) minimum to the poset of $V_4$.
$C_2$ has the following lattice
  o
  |
  o

$V_4$ has the following lattice
   o
 / | \ 
o  o  o
 \ | /
   o

$Q_8$ has the following lattice
   o
 / | \ 
o  o  o
 \ | /
   o
   |
   o

