Is a semigroup with unique right identity and left inverse a group?

group-theory semigroups

Solution 1:

As has been shown in a comment, (2) does not hold. However, (1) does.

Let $e$ be the unique right identity, and for any $x$, let $x'$ denote a left inverse.

For any $x$, $$ e = x''x' = x''ex' = x''x'xx' = ex x'. $$ Hence, for any $y$, $$ y = ye = yexx' = y xx', $$ which shows that $xx'$ is a right identity. Since it is unique, $xx' = e$. Hence every element $x$ has a two-sided inverse.

Finally, for any $x$, $$ ex = xx'x = xe = x, $$ so $e$ is a two-sided identity and the semigroup is a group.

Related

Fibonacci combinatorial identity: $F_{2n} = {n \choose 0} F_0 + {n\choose 1} F_1 + ... {n\choose n} F_n$ [duplicate]
The power set of the intersection of two sets equals the intersection of the power sets of each set
Residue of $f(z) = \frac{z}{1-\cos(z)}$ at $z=0$
Ordered binary sequences of length n with no two consecutive 0's without using recursion
Prove that a group where $a^2=e$ for all $a$ is commutative [duplicate]
solution of differential equation $\left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0$
How many 0's are at the end of 20!
Does there exist some sort of classification of finite verbally simple groups?
Determine group G as cyclic
A norm which is symmetric enough is induced by an inner product?
show that if $a | c$ and $b | c$, then $ab | c$ when $a$ is coprime to $b$.
Prove $\mathbb{R}$ is connected