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This question had two preceding parts. First, to prove that the inverse of an element $a$ is unique and second, to prove the converse of this statement. I.e. if $a$ and $b$ have inverses, then so does $a*b$. I didn't have much of a problem with both of these but this third part has me stumped.

At first, I believed, it was true and tried to work from the definition of an inverse: $(a*b)*c=c*(a*b)=e$ Using the fact that $*$ is associative, we have $a*(b*c)=e$ and $(c*a)*b=e$ which is half of the way to showing $a$ and $b$ have identities. However, I then became completely stuck on trying to show that $b*c*a=e$ (brackets omitted due to associativity) which doesn't seem to be true without commutativity.

I then assumed the statement was false and went looking for counterexamples. I tried the composition of functions and matrix multiplication as both are associative but not commutative. In the case of functions, a function, $g•f$, must be bijective to have an inverse. This combined with other theorems implies that $g$ and $f$ are bijective. Thus, the statement holds here. A similar thing arises when looking at matrix multiplication where multiplying a non-invertible matrix $A$ by any other matrix $B$ gives a non-invertible matrix $AB$. As both counterexamples failed, I am well and truly stuck and have no idea how to continue with this problem.

Thanks for any help.

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You're right, the question is false, and looking for an example with functions is a good idea. The idea below is to find an injective and a surjective function, both of which are not bijective, which composition is the identity map.

Let $X:=(\mathbb{R}^\mathbb{N})^{\mathbb{R}^\mathbb{N}}$ be the set of maps from $\mathbb{R}^\mathbb{N}$ to itself, where $\mathbb{R}^\mathbb{N}$ is the set of sequences of real numbers, and let $*$ on $X$ be the composition $\circ$. $(X,*)$ is associative and has an identity element which is the identity map \begin{align} e:\mathbb{R}^\mathbb{N}&\to \mathbb{R}^\mathbb{N}\\ x&\mapsto x. \end{align}

Define $a$ and $b$ as follows: \begin{align} a:\mathbb{R}^\mathbb{N}&\to \mathbb{R}^\mathbb{N}\\ (x_1,x_2,\dots)&\mapsto (x_2,x_3,\dots), \end{align} i.e $a$ is the map that removes the first element of the sequence, and \begin{align} b:\mathbb{R}^\mathbb{N}&\to\mathbb{R}^\mathbb{N}\\ (x_1,x_2,\dots)&\mapsto(0,x_1,x_2,\dots), \end{align} i.e, $b$ adds $0$ in the beginning of a sequence.

We have $a*b=a\circ b=e$, thus $a*b$ has an inverse. You can check that neither $a$ nor $b$ has an inverse.

Addendum: As requested by OP, I'll try to simplify the example above by giving a similar but simpler example.

Let $\mathbb{R}[X]$ be the set of polynomials with coefficients in $\mathbb R$, and let $X=\{\text{functions}\,\mathbb{R}[X]\to\mathbb{R}[X]\}$, so every element of $X$ is a function that maps polynomials to polynomials. Let $*$ be composition of functions in $X$, i.e, if $a,b\in X$, then $a*b=a\circ b$. Clearly $*$ is associative and \begin{align} e:\mathbb{R}[X]&\to\mathbb{R}[X]\\ P(X)&\mapsto P(X) \end{align} is the identity element of $(X,*)$.

We give two elements $a,b\in X$ such that $a*b=e$ yet $b*a\neq e$, which shows that $a$ and $b$ don't have inverses.

Let \begin{align} a:\mathbb{R}[X]&\to\mathbb{R}[X]\\ \sum_{i=0}^n\alpha_i X^i&\mapsto\sum_{i=1}^n\alpha_{i}X^{i-1}. \end{align}

What $a$ does is that it removes the constant term of a polynomial and reduces the degree of the other terms by $1$. One can define it by the following formula for $P(X)$ nonconstant polynomial: $$a(P(X))=\dfrac{P(X)-P(0)}{X}$$ and with $a(c(X))=0$ for constant polynomials.

Examples:

$a(5+2X+7X^2-3X^3)=2+7X-3X^2;$

$a(3-2X+9X^3)=-2+9X^2;$

$a(\sqrt{2}X-3X^3)=\sqrt{2}-3X^2.$

Now let \begin{align} b:\mathbb{R}[X]&\to\mathbb{R}[X]\\ P(X)&\mapsto XP(X). \end{align}

$b$ simply multiplies the polynomial by $X$, which has the opposite effect on the degrees compared to $a$. For example, $b(\sqrt{2}-3X^2)=\sqrt{2}X-3X^3$.

Now I claim that $a*b=e$. Indeed, if $P(X)$ is a polynomial, $b(P(X))=XP(X)$. What $b$ did is that it added $1$ degree to every term. Now what would $a(b(P(X))$ be? $a$ removes the constant term and then substracts the degree of the other terms by $1$. Since $XP(X)$ has no constant term, all $a$ does is substract by $1$ the degree of all the other terms, so it did precisely the inverse of what $b$ did, therefore $a(b(P(X))=P(X)$.

We can also easly see it as follows: for any nonzero $P(X)\in\mathbb{R}[X]$, $b(P(X))=XP(X)$ is a nonconstant polynomial, so we can use the formula: \begin{align} a(b(P(X))&=a(XP(X))\\ &=\dfrac{XP(X)-0\times P(0)}{X}\\ &=\dfrac{XP(X)}{X}\\ &=P(X). \end{align} Now I claim $b*a\neq e$. For this an example is enough: let $P(X)=1+2X+3X^2$. $a$ removes the constant term and decreases the degree of the other terms by $1$, so $a(P(X))=2+3X$. Although $b$ increases the degree by $1$, b cannot come up with the "deleted" constant term by $a$, thus $$b(a(P(X)))=b(2+3X)=2X+3X^2\neq P(X).$$

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  • $\begingroup$ Thank you very much for your answer. I'm pretty sure I understand your reasoning throughout but I am not entirely comfortable with the example. I'm sure it is correct but I am unable to visualize it very well. Are there any examples that may be slightly simpler? $\endgroup$ – Suoria Apr 14 at 8:52
  • $\begingroup$ Could I use $a(x)=lnx$, $b(x)=e^x$ in $A=\mathbf{F(R)}$ with the operation being function composition? $\endgroup$ – Suoria Apr 14 at 9:55
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    $\begingroup$ @Suoria You're welcome. I added to the answer an example on polynomials (which is in a certain sens very similar to the example I gave earlier). I invite you to read it. As for the example you suggest, the idea is not bad at all, however there's an issue: $a(x)$ is not defined on all of $\mathbb R$. $\endgroup$ – Scientifica Apr 14 at 10:01
  • $\begingroup$ Ah, of course, could you redefine $a$ to be $a(x)=ln$ when $x > 0$ and $a(x)=0$ when $x \leq 0$? $e^x$ would still not not have an inverse as it is not surjective so we still reach a contradiction. $\endgroup$ – Suoria Apr 14 at 10:07
  • $\begingroup$ @Suoria Yes that's correct! $\endgroup$ – Scientifica Apr 14 at 10:50

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