MaBloWriMo 19: groups from monoids

So, you have a monoid, that is, a set with an associative binary operation that has an identity element. But not all elements have inverses, so it is not a group. Assuming you really want a group, what can you do?

It turns out there is a very simple answer: just throw away all the elements that don’t have inverses! That is, given a monoid $M$ , form the subset $M^* \subset M$ of all the elements of $M$ that do have an inverse with respect to the binary operation. I claim that $M^*$ is in fact a group (under the same binary operation). Let’s prove it. We have to check that $M^*$ satisfies all the laws of a group.

By assumption, there is some $e \in M$ which is the identity for the binary operation. So in particular $e \odot e = e$ , which means that $e$ is its own inverse; so $e \in M^*$ .
We know all the elements in $M^*$ have an inverse in $M$ , but we still need to make sure those inverses actually end up in $M^*$ ! If $a \in M^*$ then by definition $a$ has an inverse, $a^{-1}$ , which means that $a \odot a^{-1} = a^{-1} \odot a = e$ . Note that these equations are completely symmetric: not only is $a^{-1}$ the inverse for $a$ , we can also say that $a$ is the inverse for $a^{-1}$ . So $a^{-1}$ must be in $M^*$ too.
If the binary operation is associative for $M$ then it will definitely be assocative for $M^*$ too (since associativity holds for all elements of $M$ , of which $M^*$ is a subset).
There is one more crucial thing we have to check: it is not a priori clear that $M^*$ is even closed under the binary operation—might there be some $a, b \in M^*$ such that $a \odot b$ is in $M$ but not in $M^*$ ?. But in fact if $a$ and $b$ both have inverses, then $a \odot b$ must have an inverse as well, namely, $b^{-1} \odot a^{-1}$ , since

$(b^{-1} \odot a^{-1}) \odot (a \odot b) = b^{-1} \odot (a^{-1} \odot a) \odot b = b^{-1} \odot b = e$ .

(The fact that $(a \odot b)^{-1} = b^{-1} \odot a^{-1}$ is sometimes called the “socks-shoes property”: you put on your socks first ( $a$ ), and then your shoes ( $b$ ); the inverse operation is to first take off your shoes ( $b^{-1}$ ), then your socks ( $a^{-1}$ ).)

So if $a$ and $b$ are in $M^*$ then $a \odot b$ must be as well.