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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?

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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
@Derek, Ihf: I think it's obvious that the e in the second axiom is the same as the first,but a truly "clean" axiomatic presentation would first define the right identity,the right inverse and a semigroup first.But I think the user assumes we all know what they are. –  Mathemagician1234 Sep 17 '11 at 16:16
@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
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3 Answers

up vote 6 down vote accepted

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.)

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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
@Math Thanks for the clarification and the term :). –  Srivatsan Sep 17 '11 at 16:23
@Srivatsan What I meant was that from the usual axioms of a group, you can prove the uniqueness of $e$ and $\cdot^{-1}$. –  Plop Sep 18 '11 at 10:51
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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 $a$:


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.

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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 $y:=x^{-1}x$. Then $$yy=(xx^{-1})(xx^{-1})=x(x^{-1}x)x^{-1}=x(ex^{-1})=xx^{-1}=y.$$ Hence $$e=y^{-1}y=y^{-1}(yy)=(y^{-1}y)y=(y^{-1}y)y=ey=y=xx^{-1},$$ 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
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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
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