I think the title speaks for itself. Thus I just explain the story behind the question. Of course, you may want to skip the story.

Story: Currently, I teach a course in linear algebra and matrices with a mathematician colleague of mine. In preparing for the class, we discussed together about one of the standard theorems of matrices saying the product of two invertible matrices is invertible. To understand the theorem and its converse, I naturally came to the question asked in the title. To my surprise, neither the colleague I discussed with, nor four other mathematician colleagues of mine could come up with an answer. But the question seems very natural and it is very surprising if it has been remained unnoticed. That is why I came to MO to find the answer.

PS. None of the colleagues I asked the question is an expert in algebra, and I am a mathematics educator.

PPS. You may replace "monoid" with whatever you wish, providing that you keep the rest of the title intact! Please keep your example natural (if there is one)!

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    $\begingroup$ See the first answer to this question. It also gives a counterexample to your question. $\endgroup$
    – J.-E. Pin
    Commented Sep 17, 2013 at 15:49
  • $\begingroup$ The question you linked, and its answers, indeed also included both examples and counterexamples to some of my attempts to solve the problems. Thanks. $\endgroup$ Commented Sep 17, 2013 at 18:10
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    $\begingroup$ it seems we have 3 answers with the same counterexample! $\endgroup$ Commented Sep 17, 2013 at 19:04
  • $\begingroup$ @BenjaminSteinberg And I wonder which one I shall "accept"! :) $\endgroup$ Commented Sep 17, 2013 at 19:17
  • $\begingroup$ All of them???? $\endgroup$ Commented Sep 17, 2013 at 19:19

4 Answers 4


Well, why not?

Let $\oplus_{n \in \mathbb{N}} k$ be a direct sum of countably many copies of a 1-dimensional space over a field $k$; the direct sum affords a standard basis $e_i = (0, \ldots, 0, 1, 0, \ldots)$ with $1$ in the $i^{th}$ place. Define endomorphisms $A$, $B$ by $A(e_i) = e_{i-1}$ for $i \gt 1$, $A(e_1) = e_1$, and $B(e_i) = e_{i+1}$. I think you'll agree that $A B$ (first apply $B$, then apply $A$) is the identity, but $B A$ isn't invertible.

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    $\begingroup$ Heck, why even bother with vector spaces. For monoids, just use the same idea applied to the set $\mathbb{N}$. But since this came up while discussing linear algebra... $\endgroup$
    – Todd Trimble
    Commented Sep 17, 2013 at 15:43
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    $\begingroup$ @AmirAsghari: In Todd's example, $A(e_1) = A(e_2) = e_1$, thus $A$ is not invertible. Further, $e_1 \notin {\rm im}(B)$, thus $B$ is not invertible either. $\endgroup$
    – Stefan Kohl
    Commented Sep 17, 2013 at 16:02
  • $\begingroup$ @StefanKohl At the first glance, I didn't get the point of Todd's answer. But, then I worked on the details, and I got it. Yet, I thank you for the details you added $\endgroup$ Commented Sep 17, 2013 at 18:00

There is exactly one monoid generated by two elements $p,q$ such that $pq=1$ but $qp\ne 1$. This is the bicyclic monoid $B$. In this monoid $p,q$ are not invertible while their product is 1 (hence invertible). Conversely, if a monoid contains $a,b$ such that $ab$ is invertible, then for some $c, abc=1$. If in addition $a$ is not invertible, then $bca\ne 1$. Hence $a, bc$ must generate a copy of the bicyclic monoid $B$. Thus $B$ is the smallest counterexample to your question.

  • $\begingroup$ Dear Mark. Many thanks for your answer. I "accepted" Todd's since as a "general" reader I learned more from that. $\endgroup$ Commented Sep 19, 2013 at 19:11

The shift operator $S$ and its adjoint $S^*$ on $\ell^2(\mathbb{N})$ make another example. They are defined as follows: $$S(\delta_n):=\delta_{n+1}, \quad \forall n\in \mathbb{N}$$ and $$S^*(\delta_{n+1}):=\delta_{n}, \quad \forall n\in \mathbb{N}$$ and $S^*(\delta_{1}):=0$, where $\delta_n$ is the characteristic function of the set $\{n\}$ and we extend $S$ and $S^*$ linearly. It is easy to see that $SS^*\neq id$, while $S^* S=id$. Therefore, they are not invertible, but $S^* S$ is invertible.

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    $\begingroup$ Oops, I just noticed Todd's answer uses the same idea! $\endgroup$
    – user23860
    Commented Sep 17, 2013 at 17:43
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    $\begingroup$ Dear Vahid, many thanks for your answer. I am sure it doesn't matter that much, but since your answer was essentially the same as Todd's, I accepted his answer. He was first :) $\endgroup$ Commented Sep 19, 2013 at 19:16

Since the question came from an undergraduate level perspective, here's a slight variant on the answers given thus far that students might appreciate.

Take the vector space P of real polynomials. There are two operators on P, D=differentiation and I=integration (to make integration a linear operator, we "set c=0"). Now DI is the identity map, but I is not onto (as the constant polynomials are not in its image) and D is not injective (as D(c) = 0 for any constant polynomial c).


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