Is there a set which is a group with respect to both addition and multiplication? Since addition requires 0 as it's identity and 0 has no inverse under multiplication this would seem to suggest that it is impossible but I am unable to prove it or find an example. Perhaps the rules are different enough under complex numbers, quaternions, or octonions to allow such a set to be possible.
 A: The only way this can happen if your set also has the distributive property is if the multiplicative identity equals the additive identity, i.e. $1=0$.  But this is the trivial group $\{0\}$ with $0+0=0$ and $0\cdot 0=0$.
If you do not have the distributive property, many more options are available.
A: You can just define the operations on a set to work. 
\begin{array}{c|cc}
  + & a & b \\\hline
  a & a & b \\
  b & b & a \\ 
\end{array}
\begin{array}{c|cc}
  * & a & b \\\hline
  a & b & a \\
  b & a & b \\ 
\end{array}
Here, $a$ is the additive identity and $b$ is the multiplicative identity. 
You do lose the distributive property though:
$a*(a+a)=a*(a)=b$
$a*a+a*a=b+b=a$
A: The set $\{0\}$ forms a trivial group under both of the operations $+$ and $\times$ given their ordinary meaning on the integers.
A: I think the straight forward answer is no, because a group can only have 1 operation (addition or multiplication for your case).  If you have 2 operations (addition and multiplication for your case) then it is considered a ring.
Edit (for Math Man):
"In mathematics and abstract algebra, group theory studies the algebraic structures known as groups. The concept of a group is central to abstract algebra: other well-known algebraic structures, such as rings, fields, and vector spaces, can all be seen as groups endowed with additional operations and axioms."
http://en.wikipedia.org/wiki/Group_theory
http://en.wikipedia.org/wiki/Group_%28mathematics%29
