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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?

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    $\begingroup$ At least nonemptiness needs to be required. $\endgroup$ – darij grinberg Apr 26 '19 at 13:51
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    $\begingroup$ That said, what about the binary operation $+$ defined by $a + b = a$ for all $a,b$? $\endgroup$ – darij grinberg Apr 26 '19 at 13:52
  • $\begingroup$ You have idempotens satisfying $a+a=a$ like projections (additively written), the solution to $x+a=a$ is not necessarily a neutral element $\endgroup$ – Peter Melech Apr 26 '19 at 13:54
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    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Apr 26 '19 at 15:57
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    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Apr 26 '19 at 15:58
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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.

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