Is a semigroup $G$ with left identity and right inverses a group?

Question

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?

Answer 1

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.

Answer 2

A concrete counterexample, found in John B. Frayleigh's "A First Course In Abstract Algebra", seventh edition:

Let $\Bbb R^*$ be the set of all real numbers except $0$. Define $*$ on $\Bbb R^*$ by letting $a*b$ $=$ $\lvert a \lvert b$, for all $a, b$ $\in$ $\Bbb R^*$.

Verify that this is a semigroup, contains a left identity, and a right inverse for every element in $\Bbb R^*$, but not a left inverse for every element (consider negative values), and no right identity.

Answer 3

Consider a group of any set of integers containing 1 with the product a*b = b. Then 1 is a left identity, since 1*b = b. (In fact, every number is a left identity). And 1 is a right inverse for everything, since a*1 = 1, the identity element. (Associativity is easy.) But this is obviously not a group.