Hungerford's Algebra poses the question: Is it true that a semigroup $G$ that has a left identity element and in which every element has a right inverse is a group?

Now, If both the identity and the inverse are of the same side, this is simple. For, instead of the above, say every element has a left inverse. For $a \in G$ denote this left inverse by $a^{-1}$. Then

$$(aa^{-1})(aa^{-1}) = a(a^{-1}a)a^{-1} = aa^{-1}$$

and we can use the fact that

$$cc = c \Longrightarrow c = 1$$

to get that inverses are in fact two-sided:

$$ aa^{-1} = 1$$

From which it follows that

$$a = 1 \cdot a = (aa^{-1})a = a (a^{-1}a) = a \cdot 1$$

as desired.

But in the scenario given we cannot use $cc = c \Longrightarrow c = 1$, and I can see no other way to prove this. At the same time, I cannot find a counter-example. Is there a simple resolution to this question?

  • $\begingroup$ @Behaviour flagged as duplicate of wrong question. Correct question is here Flagging again. $\endgroup$
    – Bartek
    Jan 4 '15 at 22:37
  • $\begingroup$ @Bartek But I think this one has better answers than the other one. The other one should be closed. $\endgroup$
    – user147263
    Jan 4 '15 at 22:39
  • $\begingroup$ @Behaviour Yes, but there's a third one that's already marked as duplicate of that one. I'm not sure what the convention is. I'm just letting the mods know. $\endgroup$
    – Bartek
    Jan 4 '15 at 22:41

Let $G$ have at least two elements, one of which I’ll call $e$. Define the binary operation $*$ on $G$ by $x*y=y$ for all $x,y\in G$; it’s easily checked that $*$ is associative. Clearly $e*x=x$ for all $x\in G$, so $e$ is a left identity. And $x*e=e$ for each $x\in G$, so $e$ is a right inverse for each element of $G$ (with respect to the left identity $e$). Clearly $G$ has no two-sided identity, so it isn’t a group.

Of course this is a bit odd, since I can pick any element of $G$ to be the left identity, and it then becomes the right inverse of every element.


A concrete counterexample, found in John B. Frayleigh's "A First Course In Abstract Algebra", seventh edition:

Let $\Bbb R^*$ be the set of all real numbers except $0$. Define $*$ on $\Bbb R^*$ by letting $a*b$ $=$ $\lvert a \lvert b$, for all $a, b$ $\in$ $\Bbb R^*$.

Verify that this is a semigroup, contains a left identity, and a right inverse for every element in $\Bbb R^*$, but not a left inverse for every element (consider negative values), and no right identity.

  • 1
    $\begingroup$ Here $\Bbb R^*$ has both $1$ and $-1$ as left identities. If $-1$ is the left identity when defining inverses, negative values have left inverses and the positive values don't. In general: for any choice of left identity $e$, the subset of elements which are right inverses WRT $e$ form a group with respect to the operation. Proof sketch: for any $x\in\Bbb R$ let $y$ be a right-inverse of $x$ and $z$ a right inverse of $y$. Then $e=xy=xey=x(yz)y=(xy)(zy)=ezy=zy$, i.e. $z$ is also a left inverse of $y$. Then left-identity & double-sided inverses makes the (subset of) right inverses a group. $\endgroup$
    – Frentos
    Aug 24 '17 at 12:34

Consider a group of any set of integers containing 1 with the product a*b = b. Then 1 is a left identity, since 1*b = b. (In fact, every number is a left identity). And 1 is a right inverse for everything, since a*1 = 1, the identity element. (Associativity is easy.) But this is obviously not a group.


If $cc=c$ than $c$ is called idempotent element.

Semigroup with left unit and right inverse is called left right system or shortly $(l, r)$ system.

If you take all the idempotent elements of $(l,r)$ system they also form $(l,r)$ system called idempotent $(l,r)$ system. In such system the multiplication of two elements is equal to the second element because if you take $f$ and $g$ to be two such elements, and $e$ is the unit, than


So, each element in such a system is left unit and also right inverse and any such system is not a group cause no element can be right unit.

It is now also easy to see that, if you define the multiplication of two elements to be equal to the second one, you get idempotent $(l,r)$ system, which is obviously not a group.

For more details and some extra facts check the article from 1944. by Henry B. Mann called "On certain systems which are not group". You can easily google it and find it online.

Reference to this article is mentioned somewhere at the beginning of the book "The theory of groups" by Marshall Hall.


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