# Right identity and Right inverse implies a group

Let $(G, *)$ be a semi-group. Suppose

1. $\exists e \in G$ such that $\forall a \in G,\ ae = a$;
2. $\forall a \in G, \exists a^{-1} \in G$ such that $aa^{-1} = e$.

How can we prove that $(G,*)$ is a group?

• In case you don't know: Right identity and Left inverse does not imply group. – j.p. Sep 17 '11 at 9:45
• This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"? – Derek Holt Sep 17 '11 at 15:31
• @Derek, I think the formulation is intended to be read as "Suppose there is $e\in G$ such that 1 and 2.". – lhf Sep 17 '11 at 15:52
• @lhf: Yes, that's the formally correct way to do it, and it also removes the ambiguity. – Derek Holt Sep 17 '11 at 21:29
• @ThreeFx: Take a set with two elements and define $x\cdot y=x$ (possibly $x=y$), which is associative and makes both elements to right identities. If you choose a right identity, then it's also the left inverse for both elements. – j.p. Jan 9 '17 at 8:02

I assume that (a) should read $\exists e\in G$ such that $ae=a$, $\forall a\in G$. For each $a \in G$ we have

\begin{align*} (a^{-1})^{-1}a^{-1} &= e[(a^{-1})^{-1}a^{-1}]\\ &= (aa^{-1})[(a^{-1})^{-1}a^{-1}]\\ &= [(aa^{-1})(a^{-1})^{-1}]a^{-1}\\ &= (a[a^{-1}(a^{-1})^{-1}])a^{-1}\\ &= (ae)a^{-1}\\ &= aa^{-1}. \end{align*}

Multiplying $(a^{-1})^{-1}a^{-1} = aa^{-1}$ on the right by $(a^{-1})^{-1}$ yields \begin{align*} (a^{-1})^{-1} &= (a^{-1})^{-1}e\\ &= (a^{-1})^{-1}[a^{-1}(a^{-1})^{-1}]\\ &= [(a^{-1})^{-1}a^{-1}](a^{-1})^{-1}\\ &= (aa^{-1})(a^{-1})^{-1}\\ &= a[a^{-1}(a^{-1})^{-1}]\\ &= ae\\ &= a, \end{align*}

so $a^{-1}a=e$ for all $a \in G$.

Added: The foregoing obviously assumes that $e$ is a left identity, which was not given, and somehow none of us caught it at the time. Here is a corrected argument. For each $a\in G$ we have $$a^{-1}=a^{-1}e=a^{-1}(aa^{-1})=(a^{-1}a)a^{-1}\;,$$ so $$e=a^{-1}(a^{-1})^{-1}=\left((a^{-1}a)a^{-1}\right)(a^{-1})^{-1}=(a^{-1}a)\left(a^{-1}(a^{-1})^{-1}\right)=(a^{-1}a)e=a^{-1}a\;.$$

In other words, $a^{-1}$ is both a left as well as a right inverse for $a$. It follows that

$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a\;,$$

so $e$ is a left as well as a right identity for $G$. Now you can use the usual arguments to show that the identity and inverses are unique. (For example, if $e'$ were another identity, we’d have $e = ee' = e'$, because $e$ is a left identity and $e'$ is a right identity.)

• It is not necessary to prove that $e$ and $\cdot^{-1}$ are unique, that is not part of the (usual) axioms of a group. – Plop Sep 17 '11 at 14:41
• @Plop Yes, I agree $\cdot^{-1}$ need not be unique. But if $e$ is not unique, then how do we say that $a a^{-1} = a^{-1} a = e$? (This is a doubt, not a rhetorical question :-)) – Srivatsan Sep 17 '11 at 14:49
• @Sriv: Assume e is NOT unique. Then since by definition, G is closed under multiplication, the INVERSE of each element cannot be unique either.Otherwise,we would have "isolated" identities that do not result from the product of an element and it's inverse and G would not be closed under the product!Of course,that doesn't mean we can't have an algebraic structure like this-it just means the result is not a group. There is a new concept in algebra called a Beta group,in which there are infinitely many identities and inverses,but these are not groups per se. – Mathemagician1234 Sep 17 '11 at 16:21
• In your first line $(a^{-1})^{-1}) a^{-1} = e(a^{-1})^{-1}) a^{-1}$ you assume that $e$ is a left-identity? But how you could, because it is just stated that it is a right-identity... – StefanH Nov 8 '13 at 14:41
• How is $(a^{-1})^{-1}a=e[(a^{-1})^{-1}a]$? I suppose every element has a right identity, and not necessarily a left-identity. – fierydemon Dec 2 '13 at 9:28

It is conceptually very simple that a right inverse is also a left inverse (when there is also a right identity). It follows from the axioms above in two steps:

1) Any element $a$ with the property $aa = a$ [i.e. idempotent] must be equal to the identity $e$ in the axioms, since in that case:

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

This already proves the uniqueness of the [right] identity, since any identity by definition has the property of being idempotent.

2) By the axioms, for every element $a$ there is at least one right inverse element $a^{-1}$ such that $aa^{-1}=e$. Now we form the product of the same two elements in reverse order, namely $a^{-1}a$, to see if that product also equals the identity. If so, this right inverse is also a left inverse. We only need to show that $a^{-1}a$ is idempotent, and then its equality to $e$ follows from step 1:

$$[a^{-1}a][ a^{-1}a] = a^{-1}(a a^{-1})a = a^{-1}ea = a^{-1}a$$

3) It is now clear that the right identity is also a left identity. For any $a$:

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

4) To show the uniqueness of the inverse:

Given any elements $a$ and $b$ such that $ab=e$, then

$$b = eb = a^{-1}ab = a^{-1}e = a^{-1}$$

Here, as above, the symbol $a^{-1}$ was first used to denote a representative right inverse of the element $a$. This inverse is now seen to be unique. Therefore, the symbol now signifies an operation of "inversion" which constitutes a single-valued function on the elements of the set.

See Richard A. Dean, “Elements of Abstract Algebra” (Wiley, 1967), pp 30-31.

• and where on earth did we assume idempotency in the definition of a semigroup? – moldovean Nov 6 '14 at 10:27
• @moldovean: We did not assume idempotency for any elements. It was shown that if an element is idempotent, then that element must be equal to any identity element that satisfies both 1. and 2. In a separate step, we then argued, that since any identity element (whether or not inverses exist for that element) is by definition idempotent all identity elements that exist are in fact equal to the identity element used in the proof. Ergo the identity element is unique. – Psachnodaimonia Mar 21 '15 at 12:07
• This answer gets my vote for containing the most elegant proof for the problem, that I have come across. – Psachnodaimonia Mar 21 '15 at 12:09

This is stated with left identity and left inverse as Proposition 20.4 in the book Spindler: Abstract Algebra with Applications. Let me copy here the proof from this book (it should be easy for you to change it for the right instead of left):

Let $x\in G$ be arbitrary. We want to show that the left inverse $x^{-1}$ is in fact also a right inverse. Let $y:=xx^{-1}$. Then $$yy=(xx^{-1})(xx^{-1})=x(x^{-1}x)x^{-1}=x(ex^{-1})=xx^{-1}=y.$$ Hence $$e=y^{-1}y=y^{-1}(yy)=(y^{-1}y)y=ey=y=xx^{-1},$$ i.e. $xx^{-1}=e$ which was what we wanted to show.

Now we prove that the left-neutral element $e$ is also a right-neutral element. Let $x\in G$ be arbitrary; we want to establish that $xe=x$. Now $$xe=x(x^{-1}x)=(xx^{-1})x=ex=x.$$

I googled a little and found out that several authors take this in fact as a definition of group, here are some of the first hits from google books when searching for "left inverse" "left identity" group:

• Robinson: A course in the theory of groups, p.2
• Gelbaum, Olmsted: Theorems and counterexamples in mathematics, p.1
• Sharma: Group Theory, p.14
• And of course,this is how both van der Waerden and Emil Artin define a group in their classic presentations of algebra. Later authors of textbooks generally found this "minimalist" method of defining a group far too tedious-so they assumed the stronger axioms and called it a day. A lot of classic algebra books-Herstien's TOPICS IN ALGEBRA,famously-use this as an exercise. – Mathemagician1234 Sep 17 '11 at 16:24

Most proofs I've seen contains many ingenious equations. The motivation behind these equations is not very clear. Here I present a more intuitive proof.

For any element $$a$$ in $$G$$, we can map it to a function $$f_a:G \to G$$, simply define $$f_a(x) = xa$$. We can see that every right identity element $$e$$ is mapped to the identity function $$id$$, $$f_e=id,$$ and the operation on $$G$$ amounts to function composition: $$f_{ab} = f_b \circ f_a,$$ and the existence of right inverses means for any $$a$$ there is an $$a^{-1}$$ such that $$f_{a^{-1}} \circ f_a = id.$$ That will imply that every function $$f_a$$ is injective, otherwise the composition of it, which is $$id$$, can't be injective either. Actually, it's bijective, but we don't have to use that strong conclusion here.

We want to show $$a^{-1}a=e$$. Because $$f_{a^{-1}a} = f_a \circ f_{a^{-1}}$$, we wonder whether $$f_a \circ f_{a^{-1}} =id$$. Intuitively that should be true since $$f_{a^{-1}} \circ f_a = id$$ and both $$f_a$$ and $$f_{a^{-1}}$$ are bijective. For a proof that only assumes that one of them is injective, see Lemma below.

Now that $$f_{a^{-1}a} = f_e$$, to prove $$a^{-1}$$ is a left inverse, we only need to show that for any $$c \in G$$ and $$c' \in G$$, $$f_c = f_{c'}$$ implies $$c=c'$$. We try to calculate $$f_c(e)$$: $$f_c(e) = ec = (cc^{-1})c = c(c^{-1}c)=f_{c^{-1}c}(c)=id(c)=c.$$ For the same reason $$f_{c'}(e) = c'$$, so $$c=c'$$. Note here $$ec=c$$, so it has already proved every right identity is a left identity.

The uniqueness of the identity element is induced by the uniqueness of the identity function, for $$f_e=f_{e'}$$ implies $$e=e'$$, and also the uniqueness of inverse elements is induced by the uniqueness of inverse functions.

So we've proved $$G$$ is a group.

Lemma. Let $$X$$ and $$Y$$ be two arbitrary sets. Let $$f$$ be a function $$f: X \to Y$$ and $$g$$ be another function $$g: Y \to X$$. Suppose $$g$$ is injective, and $$g \circ f = id$$, then $$f \circ g = id$$.

Proof. Let $$y \in Y$$, we want to show that $$(f \circ g)(y) = y$$. Let $$y' = (f \circ g)(y)$$. Apply $$g$$ to both side, $$g(y')=(g \circ f \circ g)(y) = (id \circ g)(y) = g(y).$$ So $$y'=y$$.