I am pretty sure I proved that the category of groups has no subobject classifier at some point. I will try and edit this post with the proof when I have more time, but I think this is an example for you to think about.

EDIT: Ya, this isn't too bad. If there was a subobject classifier $\Omega$ in Groups, then by looking at the characteristic map of the injection of $0 \rightarrow A$ for each map, you will see by writing out the diagrams that A will have to inject into $\Omega$. But then $\Omega$ is bigger than every cardinal since there is a group of every cardinality. That doesn't fly.

EDIT of EDIT: I guess I should flesh this out in greater detail.

The first thing to note is that the category of groups has a zero object (it's terminal object is also initial). I will write 0 for this.

Say Groups had a subobject classifier $0 \stackrel{true}\longrightarrow \Omega$. Note that the map true must be the unique map out of 0. Also note that the unique map $0 \stackrel{!}\rightarrow A$ is a monomorphism (i.e. injection) for every A. Thus

```
0----->0
| |
! | | true
| |
\/ \/
A ---->Ω
```

is a pullback square, where the lower map, $\chi$, is the characteristic map of !. I claim that $\chi$ is a monomorphism. This is because the ker($\chi$) maps to both A and 0 to make the diagram commute, so the inclusion of ker($\chi$) into A factors through 0 by the definition of a pullback. In other words, the kernel is trivial, so $\chi$ is an injection. Thus every group A admits an injection to $\Omega$ which is bad for set theoretic reasons.