Right identity and Right inverse implies a group

Solution 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.

Solution 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.)

Solution 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

Solution 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.