Showing uniqueness of inverse element of an element of a monoid

Question

Question- If $\langle A,*\rangle$ is a semigroup with identity, prove that every element $a \in A$ has at most one inverse.

Proof- Let the identity be $e$. Let us assume that $b_1$, $b_2$ are two distinct inverses of a belonging to $A$.

Thus $a*b_1=b_1*a=e$

Also, $a*b_2=b_2*a=e$

Now, $$\begin{array}{lrcl} & b_1 & = & b_1 \\ \implies & b_1*(a*b_1) & = & b_1*(a*b_2) \\ \implies & (b_1*a)*b_1 & = & (b_1*a)*b_2 \\ \implies & e*b_1 & = & e*b_2 \\ \implies & b_1 & = & b_2 \end{array}.$$ This contradicts the assumption that $b_1$ and $b_2$ are distinct. Thus, no element can have multiple inverses. Thus, the number of inverses is at most $1$.

Please tell me if there is any flaw in reasoning, as I can find none. Further, if my statements can be made clearer, please do comment.

Answer 1

Your proof is valid. I'm rewriting it for formatting purposes.

Proof. Let $e$ be the identity of $A$. For an element, $a\in A$, assume $b_1,b_2$ are both inverse elements of $a$. It follows that $(a*b_1)=(b_1*a)=e$ and $(b_2*a)=(a*b_2)=e$. Thus

\begin{gather} (a*b_1)=(a*b_2) \end{gather}

By left-hand multiplication we have \begin{gather} b_1(a*b_1)=b_1(a*b_2) \end{gather}

By associativity of $A$

\begin{gather} (b_1*a)b_1=(b_1*a)b_2 \implies e*b_1=e*b_2 \implies b_1=b_2 \end{gather} Therefore the inverse of an element of $A$ is unique.

Answer 2

This looks very clear to me and the proof is simple and sound. If you want to be pedantic, then the first implication maybe needs some justification using the already established fact that $a\ast b_1=e=a\ast b_2$. There is also a slightly short proof, though not by much and it depends on your taste in proof style I guess: $$b_1=b_1\ast e=b_1\ast(a\ast b_2)=(b_1\ast a)\ast b_2=e\ast b_2=b_2$$ In particular I didn't need to make an assumption that $b_1$ and $b_2$ are distinct and reach a contradiction.

Answer 3

Your proof looks good. Just one little comment: you do not really need to argue by contradiction. You can just begin by saying: "Suppose that $b_1$ and $b_2$ are inverses for $a$", and then run your argument to conclude $b_1=b_2$. This shows that there can be at most one inverse for any $a$ in a the monoid. This is primarily an aesthetic issue (your proof isn't wrong), but introducing contradiction when you can just argue directly can be a little confusing.

Answer 4

I think th word inverse in your question should be considered as the word unit. In fact, in any semigroup with identity 1, two elements for example a and b are said to be inverses of each other if we have different meaning: $$aba=a,~~~bab=b$$ and that's why any regular elemet has at least inverse in the semigroup. I know an example for a semigroup that every element are inverses to each other.