Is there a set which is a group with respect to both addition and multiplication?

Solution 1:

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.

Solution 2:

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:

$a*(a+a)=a*(a)=b$

$a*a+a*a=b+b=a$

Solution 3:

The set $\{0\}$ forms a trivial group under both of the operations $+$ and $\times$ given their ordinary meaning on the integers.

Solution 4:

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