Hungerford's Algebra poses the question: Is it true that a semigroup $G$ that has a left identity element and in which every element has a right inverse is a group?
Now, If both the identity and the inverse are of the same side, this is simple. For, instead of the above, say every element has a left inverse. For $a \in G$ denote this left inverse by $a^{-1}$. Then
$$(aa^{-1})(aa^{-1}) = a(a^{-1}a)a^{-1} = aa^{-1}$$
and we can use the fact that
$$cc = c \Longrightarrow c = 1$$
to get that inverses are in fact two-sided:
$$ aa^{-1} = 1$$
From which it follows that
$$a = 1 \cdot a = (aa^{-1})a = a (a^{-1}a) = a \cdot 1$$
as desired.
But in the scenario given we cannot use $cc = c \Longrightarrow c = 1$, and I can see no other way to prove this. At the same time, I cannot find a counter-example. Is there a simple resolution to this question?
Best Answer
Let $G$ have at least two elements, one of which I’ll call $e$. Define the binary operation $*$ on $G$ by $x*y=y$ for all $x,y\in G$; it’s easily checked that $*$ is associative. Clearly $e*x=x$ for all $x\in G$, so $e$ is a left identity. And $x*e=e$ for each $x\in G$, so $e$ is a right inverse for each element of $G$ (with respect to the left identity $e$). Clearly $G$ has no two-sided identity, so it isn’t a group.
Of course this is a bit odd, since I can pick any element of $G$ to be the left identity, and it then becomes the right inverse of every element.