# Are associativity and solvability for $x$ in all $x+a=b$) sufficient laws for a group?

Typically a group is defined as a set with a single binary operation mapping pairs of elements to elements of the set. The operation is associative, and there exists a neutral element $$0$$ such that $$0+a=a$$.

It occurs to me that if we have a set with an associative binary operation defined in it, and declare that for every equation of the form $$x+a=b$$ there exists an element $$x$$ satisfying the equation, we can show the existence of both a neutral element, as well as an inverse to every element.

That is, the solution to $$x+a=a$$ will be $$0$$. Once we demonstrate the existence of $$0$$, we require solutions to all equations $$x+a=0$$.

Are these two assumption sufficient to characterize a group?

• At least nonemptiness needs to be required. – darij grinberg Apr 26 '19 at 13:51
• That said, what about the binary operation $+$ defined by $a + b = a$ for all $a,b$? – darij grinberg Apr 26 '19 at 13:52
• You have idempotens satisfying $a+a=a$ like projections (additively written), the solution to $x+a=a$ is not necessarily a neutral element – Peter Melech Apr 26 '19 at 13:54
• First, you really shouldn't use $+$ to denote a group operation unless you mean it to be commutative, or are very explicit that you are not. Second: a solution to $x+a=a$ need not be an identity for the group, because you do not know that the solution will also be a solution to $x+b=b$. As examples, the semigroup with operation $ab=b$ for all $a,b$ satisfies your condition, but the solution(s) to the equation are not identities. – Arturo Magidin Apr 26 '19 at 15:57
• What is true is that if you have a nonempty set with a binary associative operation $*$, such that, for all $a$ and $b$, there exist solutions to both the equations $a*x=b$ and $y*a=b$, then what you have is a group. – Arturo Magidin Apr 26 '19 at 15:58

Unfortunately, it isn't, even if we add the assumption that the underlying set is non-empty. The simplest counterexample I can think of is to take any set $$S$$ with more than one element and define an operation $$*$$ on $$S$$ by $$a*b=a$$ for all $$a,b\in S.$$ Then $$a*(b*c)=a=a*c=(a*b)*c,$$ so $$*$$ is associative. Moreover, for any $$a,b\in S,$$ we have $$x=b$$ as a solution to $$x*a=b,$$ as you desired--in fact, $$x=b$$ is the unique solution! However, this operation cannot have an identity. If there were such, say $$e,$$ then taking $$a$$ and $$b$$ to be distinct elements of $$S,$$ we obtain the absurdity $$a=e*a=e=e*b=b.$$

On the other hand, let's suppose we imposed the following conditions:

• $$S$$ is a non-empty set.
• $$\star$$ is an associative operation on $$S.$$
• For any $$a,b\in S,$$ there exists some $$x\in S$$ such that $$x\star a=b.$$
• For any $$a,b\in S,$$ there exists some $$y\in S$$ such that $$a\star y=b.$$

Then $$S$$ is a group under the operation $$\star.$$

To prove it, we begin by proving that the identity exists. That is, we must show that there is some $$e\in S$$ such that for all $$a\in S,$$ $$a\star e=a=e\star a.$$

Take any $$a\in S.$$ We know that there exists $$x\in S$$ such that $$x\star a=a.$$ We'd like to show that $$x\star b=b$$ for all $$b\in S.$$ Take any $$b\in S.$$ We know there is some $$y\in S$$ such that $$a\star y=b.$$ Then $$x\star b=x\star(a\star y)=(x\star a)\star y=a\star y=b,$$ as desired. Thus, we've shown that there is at least one left-identity element in $$S.$$ A similar proof shows that there is a right-identity element in $$S.$$

Suppose that $$l,r\in S$$ are left- and right-identity elements. That is, for all $$x\in S,$$ we have $$l\star x=x=x\star r.$$ Then with $$x=r$$ we see that $$l\star r=r,$$ and with $$x=l,$$ we see that $$l\star r=l.$$ Thus, every left-identity element is a right-identity element, and vice versa, so we've proved that there is at least one identity element. We can (and should) further prove that there is exactly one identity element, but I leave that to you.

Finally, we prove the existence of inverses. Letting $$s\in S$$ be arbitrary and letting $$e$$ be the identity element, we know that there exists $$x\in S$$ such that $$x\star s=e.$$ It remains only to show that $$s\star x=e.$$ We know that there is some $$y\in S$$ such that $$(s\star x)\star y=e.$$ Since $$e$$ is the identity element, then $$s=s\star e=s\star(x\star s)=(s\star x)\star s,$$ so we see that $$s\star x=\bigl((s\star x)\star s\bigr)\star x= (s\star x)\star (s\star x),$$ and so $$\begin{eqnarray}e &=& (s\star x)\star y\\ &=& \bigl((s\star x)\star(s\star x)\bigr)\star y)\\ &=& (s\star x)\star\bigl((s\star x)\bigr)\star y)\\ &=& (s\star x)\star e\\ &=& s\star x,\end{eqnarray}$$ as desired.