A set is a set. A magma is a set with a binary operator. A semigroup is a magma with an associative binary operator. A monoid has a two-sided identity. And a group has two-sided inverses.
I am wondering about one-sided verses two-sided. Under what conditions is an identity element necessarily two-sided? Under what conditions is an inverse necessarily two-sided? And what are the simplest proofs for these?
Theorems I have so far:
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A magma may have multiple distinct left-identities or multiple distinct right-identities, but can never have a distinct left and right identity. [1: $\forall x. lx=x$. 2: $\forall x. xr=x$. 1 implies that $lr=r$ while 2 implies that $lr=l$. So either $l=r$ or at least one of 1 or 2 is false.]
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Associativity plus the existence of a two-sided inverse is enough to imply that any inverse is two-sided. [If $y$ is the left-inverse of $x$ then $xyx = x(yx)=xi=x$. By associativity, $xyx=(xy)x=x$, which implies that $xy=i$. In other words, $y$ is also the right-inverse of $x$.]
I have a feeling that an associative magma cannot have one-sided identities – but I cannot prove this.
Best Answer
The classic result in this area is that if $G$ is a semigroup with a left identity $e$ and such that every element has a left inverse with respect to $e$, then $G$ is a group. There are proofs of this all over the place, in section 1.1 of Hungerford's Algebra for example. Of course the same holds with "left" replaced with "right" throughout.
Proof: For any $g$ in $G$, we have $(gg^{-1})(gg^{-1})=g(g^{-1}g)g^{-1}=geg^{-1}=gg^{-1}$ Multiplying both sides on the left by $(gg^{-1})^{-1}$, we have $egg^{-1}=e$ and hence $gg^{-1}=e$. Thus $g^{-1}$ is in fact a two-sided inverse of $g$ (with respect to the identity $e$). Furthermore, $ge=g(g^{-1}g)=(gg^{-1})g=eg=g$, and hence $e$ is a two-sided identity.
On the counterexamples side, a fertile structure to look at is any set of at least two elements with the operation $a*b=b$. This is a semigroup in which every element is a left identity, while no element is a right identity. Furthermore, if we fix a left identity $e$, then every element has a right inverse (also $e$) with respect to $e$, while only $e$ has a left inverse (in fact everything is left inverse to it). If we relax the condition "$a$ has a left inverse" to mean "there exists $b$ such that $ba$ is a left identity" (rather than picking a specific identity and sticking to it), then everything is left inverse to everything.