mixed group

A mixed group is a partial groupoid $G$ such that $G$ contains a non-empty subset $K$, called the kernel of $G$, with the following conditions:

1. if $a,b\in G$, then $ab$ is defined iff $a\in K$,

2. if $a,b\in K$ and $c\in G$, then $(ab)c=a(bc)$,

3. if $a\in K$, then $K\subseteq aK\cap Ka$,

4. if $a\in K$ and $b\in G$ such that $ab=b$, then $ac=c$ for all $c\in G$.

Mixed groups are generalizations of groups, as the following proposition illustrates:

Proposition 1.

If $K\mathrm{=}G$, then $G$ is a group.

Proof.

$G$ is a groupoid by condition 1, and a semigroup by condition 2.

Now, by condition 3, given $a\in G$, there is $b\in G$ such that $ba=a$, so that $bc=c$ for all $c\in G$ by condition 4. In other words, $b$ is a left identity of $G$. Again, by condition 3, for every $a\in G$, there is a $d\in G$ such that $b=da$. So $ad=a(bd)=a(da)d={(ad)}^2$, so, by condition 4, $adx=x$ for all $x\in G$. In particular, set $x=a$, we get $a=(ad)a=a(da)=ab$. Hence, $b$ is a two-sided identity, and $G$ is a monoid.

Finally, by condition 3, for every $a\in G$, there are $c,d\in G$, such that $b=ac=da$. So, $c=bc=(da)c=d(ac)=db=d$, showing that $a$ has a two-sided inverse. This means that $G$ is a group.∎

For a non-trivial example of a mixed group, let $G$ be a group and $H$ a subgroup of $G$. Define a new multiplication $\cdot$ on $G$ as follows: $a\cdot b$ is defined iff $a\in H$, and if $a\cdot b$ is defined, it is defined as $ab$, the group multiplication of $a$ and $b$. Then $(G,\cdot )$ is a mixed group. Clearly, associativity of $\cdot$ is automatically satisfied. Next, pick any $a\in H$, then, for any $b\in H$, $a^{-1}\cdot b$ and $b\cdot a^{-1}$ are both elements of $H$, so that $b\in a\cdot H\cap H\cdot a$, and condition 3 is also satisfied. Finally, if $a\in H$ and $b\in G$ such that $a\cdot b=b$, then $a$ is the multiplicative identity of $G$, clearly $a\cdot c=c$ for all $c\in G$.