I have some difficulties with a task in algebra. I guess it's trivial and really easy but I can't figure out how to solve it.

I have a set $G$ and a binary operation on it, let it be $\circ$. I have that the operation is associative and that the equations $a\circ x = b$ and $x\circ a = b$ have unique solutions. I have to prove that $(G, \circ)$ is a group.

I already have that the operation is binary and associative, so I have to prove that there is unique identity element and unique inverse element and it will come from the equations, but how exactly?

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    $\begingroup$ I've edited your question to use $\LaTeX$. Please make sure it still represents your original intent. For help with formatting in the future, please see this meta question. $\endgroup$
    – apnorton
    Commented Mar 30, 2013 at 16:25
  • $\begingroup$ Yes, it still represents the original content! Thank you very much for the edition and the link! $\endgroup$
    – Faery
    Commented Mar 30, 2013 at 16:27
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    $\begingroup$ Just a quick note, you only have to show that there exists some element that acts as an identity for the whole set, and some element that acts as an inverse, for a given element. The uniqueness of these things is a consequence of the group axioms, not inherent within them. $\endgroup$ Commented Mar 30, 2013 at 16:33
  • $\begingroup$ This question requires one more assumption: that the set $G$ is non-empty. All the given answers implicitly assume this, and necessarily so, since $G = \emptyset$ satisfies all your current hypotheses but is not a group. $\endgroup$ Commented Oct 4, 2013 at 16:51

4 Answers 4


Just for the heck of it, we can do it even without the uniqueness assumption.

Assumed: that multiplication is associative and that the equations $ax=b$ and $xa=b$ have at least one solution for all $a$ and $b$.

  1. If $e$ and $a$ are given such that $ea=a$, then $e$ is a left identify for every element. Proof: Given $b$ let $x$ be such that $ax=b$. Then $eb=eax=ax=b$.

  2. Similarly, anything that is a right identity for something is a right identity for everything.

  3. If $l$ is a left identity for something and $r$ is a right identity for something, then $l=r$. Proof: By (1) and (2) $lr=l$ and $lr=r$.

  4. There is a unique identity element. Namely pick any element $a$ and let $e$ solve $ea=a$. Then by (3) the solution to $ax=a$ must equal this $e$, and then by (3) again, everything that is a right or left identity of something must be $e$.

  5. Uniqueness of inverses now follows by the usual argument: If $xa=ya=e$, then let $z$ be such that $az=e$ and then $x=xe=xaz=yaz=ye=y$ as well as $z=ez=xaz=xe=x$.

Come to think of it, this is actually a nicer characterization of groups than the usual mucking around with identities and inverses. A group is simply something with an associative operation such that everything can be made into anything else by multiplying it with something form the left or from the right, as we choose. Borrowing some terminology from graph theory we might call this a left-transitive and right-transitive associative operation. (Edit: It turns out this already has a name, though: Left simple semigroup and right simple semigroup -- and what the above proves is that groups are exactly those semigroups that are both left and right simple).

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    $\begingroup$ One small note — for this, as for the original question, we need one more assumption: $G$ should be non-empty! $G = \emptyset$ is consistent with all the original hypotheses. $\endgroup$ Commented Oct 4, 2013 at 16:49

Hint First, note that from uniqueness of solutions it follows that the operation is cancellative: $ac=bc$ (as well as $ca=cb$) implies $a=b$.

The equation $ax = a$ has a solution $e$. Multiplying the identity $ae=a$ on the right by $a$ and then cancelling $a$ on the left we get $ea=a$, i.e. $e$ is an unity for $a$. Multiplying and cancelling by other element $b$ we prove that $e$ is an unity for all $G$. Next, from the equations $ax = e$ and $xa = e$ we obtain inverse elements...

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    $\begingroup$ Thank you very much! Now I see the meaning of the equations have unique solutions for each a and b from G in a different way! $\endgroup$
    – Faery
    Commented Mar 30, 2013 at 16:30
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    $\begingroup$ @Boris I think the hard part about this problem is showing that $x$ will be the same for every $a$. I don't see how that follows (only then you can conclude that there exists an identity). $\endgroup$
    – Git Gud
    Commented Mar 30, 2013 at 16:31
  • $\begingroup$ @Git Gud: Certainly. But I only wrote a hint! It seems it is enough for Faery. $\endgroup$ Commented Mar 30, 2013 at 16:35
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    $\begingroup$ I think it's more likely that he has accepted your answer without understanding the subtlety that Git Gud alluded to. The solution to the problem is not nearly as simple as your hint suggests. $\endgroup$ Commented Mar 30, 2013 at 16:36
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    $\begingroup$ However, if $ea=a$ and $ax=b$, then $eb=eax=ax=b$. $\endgroup$ Commented Mar 30, 2013 at 16:43

It is a bit subtle to prove the existence of an identity element. For each $g \in G$, there exist unique elements $l_g, r_g$ (left/right identities for $g$) such that $l_g g = g$ and $g r_g = g$. The goal is to show that $l_g = r_g = l_{g'} = r_{g'}$ for any two $g, g' \in G$.

You have $g l_g g = g^2$ and since the equation $x g = g^2$ has a unique solution, we have $g l_g = g$. Thus, using uniqueness again, $l_g = r_g$. Let's call this element $e_g$, which has the property that $e_g g = g e_g = g$.

Now we must show that $e_g = e_{g'}$ for $g, g' \in G$. We have $g e_g g' = g g' = g e_{g'} g'$. Since the equation $x g' = g g'$ has a unique solution, $g e_g = g e_{g'}$; similarly, since $g x = g e_{g}$ has a unique solution, $e_g = e_{g'}$.


Let $g\in G$. By hypothesis, you know that the equation $gx=g$ has a unique solution and $xg=g$ has a unique solution. What can you say about these two soltuions? Let $x_0$ be the solution of the first equation. By hypothesis, the equations $gx=x_0$ and $xg=x_0$ have unique solutions. What can you say about these two solutions?

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    $\begingroup$ "This $x$" is really two, a priori possibly different, $x$'s. $\endgroup$ Commented Mar 30, 2013 at 16:43
  • $\begingroup$ Editted to reflect that you must show at some point that the solutions to both equations are equal. $\endgroup$
    – Dan Rust
    Commented Mar 30, 2013 at 16:45

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