# Probs. 12 & 13 , Sec. 2.3, in Herstein's TOPICS IN ALGEBRA, 2nd ed: Existence of only right-sided identity and right-sided inverses suffice

I. Let $$G$$ be a nonempty set closed under an associative product, which in addition satisfies:

(a) There exists an $$e\in G$$ such that $$a\cdot e=a$$ for all $$a\in G$$.

(b) Give $$a\in G$$, there exists an element $$a^{-1}\in G$$ such that $$a\cdot a^{-1}=e$$.

Prove that $$G$$ must be a group under this product.

II. Prove, by an example, that right indentity element and left inverse does not imply that $$G$$ is group.

My solution:

I. Since $$G$$ is closed set under an associative product, i.e. if $$a,b,c\in G$$ then $$(a\cdot b)\cdot c=a\cdot (b\cdot c)\in G$$. Taking $$c=e$$ we get $$(a\cdot b)\cdot e=a\cdot (b\cdot e)=a\cdot b \in G$$. We have shown that $$\cdot$$ is binary operation.

Since $$a\in G$$ then $$a^{-1}\in G$$ and we have the following identities $$a^{-1}=a^{-1}\cdot e=a^{-1}\cdot (a\cdot a^{-1})=(a^{-1}\cdot a)\cdot a^{-1}$$ Then \begin{align} e &= a^{-1}\cdot (a^{-1})^{-1} \\ &= \left( \left( a^{-1}\cdot a \right) \cdot a^{-1} \right) \cdot \left( a^{-1}\right)^{-1} \\ &= \left( a^{-1}\cdot a \right) \cdot \left( a^{-1}\cdot \left( a^{-1} \right)^{-1} \right) \\ &= \left( a^{-1}\cdot a \right) \cdot e \\ &=a^{-1}\cdot a. \end{align} Thus we have shown that $$a\cdot a^{-1}=a^{-1}\cdot a=e.$$ Then we see that $$e\cdot a= \left( a\cdot a^{-1} \right) \cdot a = a \cdot \left( a^{-1} \cdot a \right) = a \cdot e = a.$$

We have shown that for this set $$G$$ and the associative binary operation assumed to be defined on $$G$$, the properties of the existence of a two-sided identity element in $$G$$ and the existence in $$G$$ of a two-sided inverse for each element of $$G$$ are satisfied. Therefore $$G$$ is indeed a group.

II. But II indeed is true. Lets take the set $$G=\{a,b,e\}$$ and define the product $$\cdot$$ by the following identities: $$e\cdot e=a\cdot e=b\cdot e=e$$ and $$a^{-1}=b, \ b^{-1}=a$$ and consider the following multiplication table for our set $$G$$

$$\begin{array}{c | c c c c c} \hline\hline & e & a & b \\ \hline e & e & b & b & \\ a & a & a & e & \\ b & b & e & a & \\ \hline \end{array}$$

It's easy to verify that conditions of second problem hold for our $$G$$, however, $$G$$ is not group since we can show that $$b=a$$.

Is my reasoning above correct?

EDIT: Maybe this is a duplicate but I would like to know if my solution is true since I have solved it by myself. Especiaaly I am interested in the solution of the second problem.

• Maybe it is a duplicate but I would like to know if my solution is true? – ZFR Nov 22 '17 at 8:58
• Part I is correct. – Nex Nov 22 '17 at 19:43
• @Nex, What about part II? – ZFR Nov 22 '17 at 20:04
• Your construction seems to be ad hock and hence time consuming to check if it is associative or not. Why not $x\cdot y= x$ on the same set? – Nex Nov 22 '17 at 20:32
• @ZFR I've made some edits to your post. Do you approve of this and agree to my amendments? – Saaqib Mahmood Oct 22 '19 at 7:44

You don't need to verify that the operation is defined and associative: that's already given.

What you need to show is that

1. $$e$$ is a left identity as well as a right identity (the latter condition is given)
2. $$a^{-1}a=e$$, for every $$a\in G$$

On the other hand, using $$a^{-1}$$ may be misleading, but your argument seems good. For the sake of clarity, I'll denote by $$b$$ and $$c$$ elements such that $$ab=e$$ and $$bc=e$$. Your argument becomes $$b=be=b(ab)=(ba)b$$ then $$e=bc=((ba)b)c=(ba)(bc)=(ba)e=ba$$ Therefore $$ea=(ab)a=a(ba)=ae=a$$ Good work!

The operation you give the Cayley table of does not define a group structure on $$\{e,a,b\}$$, because $$ea=eb$$, but $$a\ne b$$ (this is better than saying that “we can show that $$a=b$$, which is false at the outset). So long as you verify it is associative, you have your counterexample.

I think your proofs of the right identity also being a left identity and the right-inverse of an element also being a left-inverse are OK.

However, talking about closedness (or associativity) is overkill since these conditions are already assumed.

As for a counter-example, how about the following example?

Let us choose $$G$$ to be the set of all non-zero real numbers, and for any elements $$a, b \in G$$, let us define $$a*b$$ as follows: $$a*b \colon= a \, \lvert b \rvert.$$

Then we note that, for any element $$a \in G$$, we have $$a * 1 = a = a * (-1).$$ Thus both $$1$$ and $$-1$$ act as our right identity elements.

And, for each element $$a \in G$$, we note that the element $$\frac{1}{\lvert a \rvert } \in G$$ and we obtain $$\frac{1}{ \lvert a \rvert } * a = \frac{1}{ \lvert a \rvert }\, \lvert a \rvert = 1,$$ one of the two identity elements.

Also for each $$a \in G$$, the element $$-\frac{1}{\lvert a \rvert } \in G$$ satisfies $$-\frac{1}{\lvert a \rvert} * a = -1,$$ the other one of the two right identity elements.

As for associativity, for any elements $$a, b, c \in G$$, we note that $$a*(b*c) = a * \big(b \, \lvert c \rvert \big) = a \, \big\lvert b \, \lvert c \rvert \big\rvert = a \, \lvert bc \rvert = a \, \lvert b \rvert \, \lvert c \rvert = \big( a \, \lvert b \rvert \big) \, \lvert c \rvert = (a*b) \, \lvert c \rvert = (a*b)*c.$$

However, this set $$G$$ with this binary operation $$*$$ is not a group. Because for any $$a \in G$$, we note that $$1 * a = 1 \, \lvert a \rvert = \lvert a \rvert \neq a$$ when $$a < 0$$, and $$(-1) * a = (-1) \, \lvert a \rvert = - \lvert a \rvert \neq a$$ when $$a > 0$$. Morever, if there were an element $$e \in G$$ such that $$e*a = a,$$ then that element $$e$$ would satisfy $$e \, \lvert a \rvert = a,$$ and hence $$e = \frac{a}{\lvert a \rvert } = \pm 1.$$ This shows that $$G$$ has no left identity element. Therefore $$G$$ cannot be a group.

Hope this helps.

• @ZFR please have a look at my answer. – Saaqib Mahmood Oct 22 '19 at 7:45