# group theory abstract algebra

• Oct 23rd 2009, 03:06 PM
bettywhit
group theory abstract algebra
consider all the rationals except 1 with the binary operation a*b=a+b-ab for any a,b elements in Q\ {1}.
show that this is a group.
why is 1 excluded
• Oct 23rd 2009, 03:39 PM
proscientia
Verify that $\displaystyle \left\langle\mathbb Q\setminus\{1\},\star\right\rangle$ satisfies all the group axioms. (Do you know what the group axioms are?) Hint for identity: $\displaystyle a\star0=0\star a=\ldots\,?$

This group is isomorphic to $\displaystyle \left\langle\mathbb Q\setminus\{0\},\times\right\rangle$ (the group of nonzero rationals under ordinary multiplication) via the mapping $\displaystyle a\mapsto\frac1{1-a}$ for $\displaystyle a\in\mathbb Q\setminus\{1\}.$
• Oct 25th 2009, 03:04 PM
bettywhit
Quote:

Originally Posted by bettywhit
consider all the rationals except 1 with the binary operation a*b=a+b-ab for any a,b elements in Q\ {1}.
show that this is a group.
why is 1 excluded

thanks i figured out how to show it is a group but i am not sure why 1 is excluded
• Oct 25th 2009, 03:23 PM
tonio
Quote:

Originally Posted by bettywhit
thanks i figured out how to show it is a group but i am not sure why 1 is excluded

The group's unit is 0, since $\displaystyle a*0=a+0-a \cdot 0=a$. If 1 were an element of this group, then you'd get $\displaystyle a*1=a+1-a\cdot 1=a+1-1=a$ and you have ANOTHER unit, which cannot be in a group.

Tonio
• Oct 26th 2009, 04:41 PM
proscientia
Something is wrong here.

$\displaystyle a\star1\ =\ a+1-a\cdot1=a+1-\color{red}a\color{black}=\color{red}1$

You just cannot have two identities, even in a semigroup! Including $\displaystyle 1$ would make $\displaystyle \left\langle\mathbb Q,\star\right\rangle$ a semigroup rather than a group (the element $\displaystyle 1$ will not have an inverse). If a semigroup has an identity, then that identity is always unique.
• Oct 26th 2009, 04:54 PM
sfspitfire23
If 1 was included, the inverse would also be undefined,