Binary operation ab+a defined on Q. Is it a group?

Question

A binary operation ∗ is defined on Q such that a∗b=ab+a. Is it a group? I think that it's associative and the identity element is 0, but what about the inverse element? Am I right to think that for all a it's -1?

Answer 1

This is not a group; there is a right identity, which is $0$: $$a*0 = a0 + a = a$$ and every element has a right inverse, which is $-1$ for all $a$ $$a*(-1) = a(-1)+a = 0$$ but the fact that the same element works as a right inverse for everyone already tells you this cannot possibly be a group (if $x^{-1}=y^{-1}$ in a group, then $x=y$).

In fact, the operation is not associative, since for example: $$(3*2)*1 = (6+3)+9 = 18$$ but $$3*(2*1) = 3(2+2)+3 = 15$$ (among many other possibilities).

Moreover, the right identity is not a left identity, since $0*a = 0a+0 = 0$, which will not equal $a$ for all $a\neq 0$; and if you are given $a$, then $x*a = 0$ requires $0=xa+x = x(a+1)$, so only $a=-1$ has a left inverse.

So this is a magma with a right identity and right inverses, but not group.

Answer 2

The binary operation is not associative. $a*(b*c)=a*(bc+b)=a(bc+b)+a=abc+ab+a\\ (a*b)*c=(ab+a)*c=(ab+a)c+(ab+a)=abc+ac+ab+a\\$ which need not be equal in general. For example, take $a=b=c=1$ then $a*(b*c)=3$ and $(a*b)*c=4$.

Answer 3

It is not a group, since$$a\ast a=a\iff a^2+a=a\iff a=0.$$So, only $0$ can be the identity element. But $0*1=0\ne1$. So, $(\Bbb Q,\ast)$ has no identity element.

Answer 4

It doesn't look associative to me and $a*(b*c) = a*(bc +b)= abc + ab + a$ and $(a*b)*c = (ab + a)*c= (ab+a)c + (ab+a) =abc + ac +ab +a$ which are of course not always equal.

To find an identity we must find an $x$ so that 1) $a*x=ax + a= a$ always and 2)$x*a= x*a +x = a$ always.

The so solution to 1) is $ax = 0$ so $x = 0$ for all $a$. But the solution to 2) is $x(a+1)=a$ and if $a\ne 1$ then $x = \frac a{a+1}$ which is not a consistent value for every $a$. So there is a right-identity but not an identity.

To find a right inverse to the right identity (There's no point in finding a left inverse to a right identity because we could never use that to solve anything--- although as $*$ isn't associative there's not point in finding an inverse at all because we can only use inverses to solve things when $*$ is associative) we must solve $a*x = ax + a=0$ so $x=-1$. Which is interesting be we can use it to solve $ay = c$ because although $(ay)*-1 = c*(-1)$ is true and $a(y*-1)= a*0 = a$ is true and is not true that $(ay)*-1= a(y*-1)$, and also $c*(-1) = 0$ so this will never solve anything.

So no, it isn't a group.

Oh, what the heck to find a left inverse to the right identity.... $x*a =0$ means $xa+x=0$ and $x(a+1)=0$ so $x=0$.