Semigroup could be a group?

Question

Is it true that a semigroup which has a left identity element and in which every element has a right inverse is a group?

My attempt: Let $G$ a semigroup and $a\in G$, so exist $a^{-1}\in G$ such that $a*a^{-1}=e$. Also, $a*a^{-1}\in G$. Then: $(a^{-1}*a)(a^{-1}*a)^{-1}=(a^{-1}*a)(a^{-1}*a)=a^{-1}*(a*a^{-1})*a=a^{-1}*e*a=a^{-1}*a=e$. So, $a$ has leftt inverse, therefore $G$ is group.

Is correct my attempt? Something wrong? I really appreciate your help.

Answer 1

Your argument is necessarily wrong, here's why.

Call $S\neq \emptyset$ a right zero semigroup if $xy = y$ for all $x, y\in S$. Pick any $e\in S$. Then $e$ is a left identity of $S$, and for any $x\in S$ there is $x^{-1}\in S$ such that $xx^{-1} = e$, namely we can set $x^{-1} = e$.

In fact, semigroups which have left identity and right inverses (wrt this left identity) are precisely those isomorphic to one of the form $G\times S$ for some right zero semigroup $S$ and a group $G$, where $G\times S$ is a product of semigroups.

For such a semigroup to be a group you need either both left identity and left inverses, or both right identity and right inverses.

Answer 2

There are multiple errors in your derivation. I will avoid the use of ${}^{-1}$, since that suggests two-sided-ness.

Let $b$ be a right inverse of $a$, so that $ab=e$; we also know that $ex=x$ for all $x$.

1. You claim that a right inverse of $ba$ is $ba$. You seem to be using two facts, borrowed from groups: that $(xy)^{-1}=y^{-1}x^{-1}$, and that $(x^{-1})^{-1}=x$. But neither of these equalities is justified here. Note that the right inverse of $ba$ is $ba$ if and only if $b$ is a two-sided inverse of $a$ (given that it is already a right inverse). Indeed, we always have $(ba)(ba) = b((ab)a) = b(ea) = ba$ if $b$ is a two-sided inverse, then $(ba)(ba) = ba = e$; so $(ba)$ is its own inverse; and if $(ba)(ba)=e$ then that means $ba=e$, so $b$ is a two-sided inverse of $a$. Thus, your first equality (going from $(a^{-1}*a)^{-1}$ to $(a^{-1}*a)$) already assumes that you have a two-sided inverse and hence that $a$ has a left inverse... but that is what you are attempting to establish. It's a circular argument.

2. You then claim at the end that $ba=e$, which is again the same assumption.

Note that if $b$ is a right inverse of $a$, and $y$ is a right inverse of $x$, then it is of course true that $yb$ is a right inverse of $ax$: for $$(ax)(yb) = a((xy)b) = a(eb) = ab = e.$$ That means that you could replace, in your notation, $(a^{-1}*a)^{-1}$ with $a^{-1}*(a^{-1})^{-1}$. But here you are tacitly assuming that $(a^{-1})^{-1}=a$... and again that amounts to assuming that if $ab=e$, then $ba=e$... which is what you are trying to prove.

So the reason your argument is flawed is because in attempting to prove that if $b$ is a right inverse of $a$ then it is also a left inverse of $a$, you invoke numerous times results that are equivalent to assuming that $b$ is a two-sided inverse of $a$; your argument is entirely circular.