Right identity and Right inverse in a semigroup imply it is a group

Question

Let $(G, *)$ be a semigroup. 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?

Answer 1

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.

Answer 2

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.)

Answer 3

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

Answer 4

Martin's answer, but using a notation that may be easier to follow.

Let $G = (S,\cdot)$ be a semi-group such that it has a right identity $1_{r}$, and that each of its elements has a right inverse; i.e.,

1. $\forall x \in G: x1_{r} = x$
2. $\forall x \in G: \exists x^{-1}_{r}: xx^{-1}_{r} = 1_{r}$

We show that $G$ is in fact a group:

Let $y = x^{-1}_{r}x$.

$$yy = (x^{-1}_{r}x)(x_{r}^{-1}x) = x^{-1}_{r}(xx^{-1}_{r})x = x^{-1}_{r}(1_{r})x = (x^{-1}_{r}1_{r})x = x^{-1}_{r}x$$

Therefore, $y$ is idempotent, and we can use that to show that $x^{-1}$ is actually a two-sided inverse:

$$1_{r} = yy^{-1}_{r} = y(yy^{-1}_{r}) = y1_{r} = y = x^{-1}_{r}x$$

Then, that $1_{r}$ is also two-sided:

$$1_{r}x = (xx^{-1}_{r})x = x(x^{-1}_{r}x) = x1_{r} = x$$

This suffices to show that $G$ is a group.

Bonus

This allows us to quickly demonstrate that any semi-group in which $ax = b$ and $ya = b$ are uniquely determined is also a group:

Let $G = (S, \cdot)$ be a semi-group with such proprieties, and $a$ be an element of that group. We know $ax = a$ must have a unique solution, lets call it $e$. Then for any $b \in G$:

$$be = yae = ya = b$$

Where the first equality is also due to $ya = b$ being uniquely determined. We now have everything we need: $e$ is a right identity, and each element of $G$ has an inverse.

Answer 5

This is discussed on page 7 of Serge Lang's Algebra (2002). He proves an analogous statement about a left identity and left inverses, so I adapt his proof for the question at hand.

Definition. A semigroup is a set together with an associative binary operation.

Theorem. Suppose a semigroup $G$ contains an element $e$ with the following two properties: (i) $ae=a$ for all $a\in G$, and (ii) for all $a\in G$ there is a $b\in G$ such that $ab=e$. Then, $G$ is a group, and $e$ is its (two-sided) identity element.

Proof. Let $x\in G$, and let $y\in G$ be a right inverse of $x$ (i.e. let $y$ be an element of $G$ satisfying $xy=e$). Note that $yxy=y$. If we multiply both sides of this equation on the right by a right inverse of $y$, we obtain $yx=e$, and so $y$ is also a left inverse of $x$. It now follows that $e$ is a left identity, since $ex=xyx=xe=x$.

Answer 6

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 $y'=y$.

Answer 7

Here are two more proofs that I think are different to the ones already given. Although I am probably biased, the first proof at least seems more direct than the ones given already in that these either proceed using idempotents or via the action of the group on itself.

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Notation etc.: A pair $(S,*)$ consisting of a set $S$ and an associative binary operation $*\colon S \times S \to S$ is called a semigroup. An element $e\in S$ is a left/right/2-sided identity if L1, R1 respectively T1 holds, that is, for all $s \in S$ we have

• L1/R1/T1: $e*s = s$, resp. $s*e=s$, resp. $e*s=s*e =s$.

If $(S,*,e)$ is a semigroup $(S,*)$ together with an identity $e$ (which may be left, right or 2-sided, then $(S,*,e)$ has left/right/2-sided inverses if

• L2/R2/T2: for every $a \in S$ there is some $b\in S$ with $b*a=e$, resp. $a*b=e$, resp $a*b=b*a =e$.

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Proposition: A semigroup $(S,*)$ with a left identity $e$ satisfying L2 is a group.

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Proof 1: For any $s \in S$, by L2, we can find some $t \in S$ with $t*s=e$. Applying L2 to $t \in S$ we can find an $s' \in S$ with $s'*t = e$, so that $t$ has both a left and a right inverse. Moreover, we have $$ \tag{$\dagger$} s=e*s = (s'*t)*s = s'*(t*s) = s'*e. $$ But now $s*t = (s'*e)*t= s'*(e*t) = s'*t = e$, so that $s$ is a 2-sided inverse for $t$, or equivalently $t$ is a 2-sided inverse for $s$.

But now to see that $e$ is a 2-sided identity note that if $s \in S$ has $t$ as a 2-sided identity then we have $s=e*s = (s*t)*s= s*(t*s) = s*e$.

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Proof 2: Let $S*e :=\{s*e: s \in S\}$. We first claim that $e$ is a 2-sided identity if and only if $S = S*e$. Indeed if $e$ is a 2-sided identity then clearly $S = S*e$. Conversely, if $s*e \in S*e$ then $(s*e)*e = s*(e*e) = s*e$, so that $e$ is a 2-sided identity on $S*e$.

Now if $s\in S$ by L2 it has a left inverse $t$ which itself has a left-inverse $s'$. But then $$ \tag{$\dagger$} s = e*s = (s'*t)*s = s'*(t*s) = s'*e, $$ so that $s \in S*e$. Since $s \in S$ was arbitrary we must have $S= S*e$, and hence $e$ is a 2-sided identity. But then $(\dagger)$ becomes $s = s'$, hence $s$ is a 2-sided inverse for $t$, i.e. $t$ is a 2-sided inverse for $s$, and since $s$ was arbitrary, $S$ is a group as required.

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Comment: Although both proofs rely fundamentally on ($\dagger$), their logic appears to me to be genuinely distinct -- indeed proof 1 shows that 2-sided inverses exist for $e$ and deduces from this that $e$ is a 2-sided identity, while in the second proof we use the fact that $e$ is a 2-sided identity to deduce that $s$ has a 2-sided inverse.

Answer 8

Thinking of the elements of a semigroup as functions makes the proof of this fact much easier to understand.

For $a \in G$, we can define a function $\sigma_a : x \mapsto x * a$. By the existence of right inverses, we have: \begin{align*} \sigma_{a^{-1}} &\circ \sigma_a = \operatorname{id}_G \\ \sigma_{(a^{-1})^{-1}} &\circ \sigma_{a^{-1}} = \operatorname{id}_G \end{align*} This shows that $\sigma_{a^{-1}}$ has both left and right inverses, so both $\sigma_{a^{-1}}$ and its inverse $\sigma_{a}$ are bijective.

Having showed that $\sigma_a$ is bijective, it is very easy to deduce that $e$ is also a left identity. There exists an $x \in G$ such that $y * a = a$. Canceling the $a$'s by multiplying with $a^{-1}$ from right, we conclude that $x = e$, so $e * a = a$ for all $a \in G$.

Moreover, showing that right inverses are also left inverse is easy as well. There exists a $y \in G$ such that $x * a = e$. Again by multiplying with $a^{-1}$ from right, we get $x = a^{-1}$, so $a^{-1} a = e$ for all $a \in G$.

In summary, $e$ is actually two sided identity, and $a^{-1}$ is actually two sided inverse of $a$.