# Show that $a \star b=a \cdot b+a+b$ is binary operation for the group $\Bbb{ Q} \setminus \{-1\}$

The group $$\left(\Bbb{ Q}\setminus\{-1\},\star\right)$$ has as its underlying set the rational numbers different from $$-1$$ and the operation $$\star$$ is defined as $$a \star b=a \cdot b+a+b$$ where multiplication/addition are the usual operations with rational numbers.Show that this is a binary operation.

identity element has to be 0 since $$a\star 0 = a+0+a(0)$$ so for $$a\star b=0$$ we have $$ab+a+b=0$$ $$ab+a=-b \\ a(b+1)=-b \\ a=\frac{-b}{b+1}$$ so $$b$$ cant be $$-1$$ since that solution is not in $$\Bbb{Q}$$. Similar question might have been answered here. Is this operation true for all groups? I think so.

Thanks for the tips and help.

Your solution is correct (though I might try plugging in your inverse value - just for formality, it's more correct to go that way) - you could do it a little more elegantly by noting that, if we let $f(x)=x+1$, then the operation happens to be $x*y=f^{-1}(f(x)\cdot f(y))=(x+1)(y+1)-1$. So, basically, what this says is that $x*y$ is just multiplication, except that we "project" each argument to $x+1$ before the operation, and then undo it later. Notice that the identity of this group, $0$, has that $f(0)=1$ - the identity of $(\mathbb{Q}-\{0\},\cdot)$, and that the inverse of $x$ in your group is $f^{-1}(\frac{1}{f(x)})$ - check that by computation, if you wish.
We could do this for any bijective function $f$, not just $f(x)=x+1$. So operations like $x*y=2xy$ - where $f(x)=2x$ - or $x*y = xy-x-y+2$, where $f(x)=x-1$ are also groups.
More generally, let $f:S\rightarrow G$ be a bijection, where $S$ is a set and $(G,\cdot)$ is a group, with identity $e$ and inverses $x^{-1}$ for any $x$. Then, we can prove that, for $x,y\in S$ the operation $$x*y=f^{-1}(f(x)\cdot f(y))$$ form a group $(S,*)$. Note that it has an identity $f^{-1}(e)$, since $$x*f^{-1}(e)=f^{-1}(f(x)\cdot f(f^{-1}(e)))=f^{-1}(f(x)\cdot e)=f^{-1}(f(x))=x$$ and a proof of a similar flavor can show associativity and that inverses exist, since $x*f^{-1}(f(x)^{-1})$ is the identity. This, in particular proves that, so long as $f(x)=x+1$ is bijective, which it ought to be in any ring (a generalization of a group, including two operations - which is probably what you mean), then the new operation is a group. In a group, for any element $a$, the map $x*y=a^{-1}(ax)\cdot(ay)=xay$ always forms a group as well.
• Well, if it's a bijection, yeah. So $f(x)=mx+b$ always works, and yields the operation $x*y=((mx+b)(my+b)-b)/m$ which expands to $x*y=mxy+bx+by+\frac{1}m(b^2-b)$. Commented Oct 19, 2014 at 3:31