Algebra: operations Is there a multiplicative identity with additive inverse of infinite multiplicative order? In all the examples I know, the additive inverse of the multiplicative identity has multiplicative order 2. For example, the additive inverse of the multiplicative identity 1 is -1 and the multiplicative order of -1 is 2. I need a counter-example involving two familiar operations, not necessarily multiplication and addition.    
 A: If you're really asking about arbitrary operations, as your comments suggest, then this is easy:


*

*We'll define two operations $\oplus$ and $\otimes$ on $\mathbb{N}$.

*Our first operation, $\otimes$, is just the usual multiplication. So the $\otimes$-identity is 1.

*Our second operation, $\oplus$, is given by setting $x\oplus 2=2\oplus x=x$ (so $2$ is the $\oplus$-identity), $1\oplus 3=3\oplus 1=2$, and - if $x, y\not=2$ and $\{x, y\}\not=\{1, 3\}$ - then $x\oplus y=7$. This is a very stupid operation.
Now note that $1$ has a $\oplus$-inverse, namely $3$. But clearly $3$ has infinite $\otimes$-order.
This should be a cautionary tale: practically any behavior you want can occur in some algebraic structure, it's usually when we ask for certain phenomena to occur while obeying global rules (e.g. associativity, commutativity, . . .) that things get iteresting. In particular, when you have multiple operations under consideration, they will usually be related in some way (e.g., distributivity).
