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.

  • $\begingroup$ Maybe it is a duplicate but I would like to know if my solution is true? $\endgroup$ – ZFR Nov 22 '17 at 8:58
  • $\begingroup$ Part I is correct. $\endgroup$ – Nex Nov 22 '17 at 19:43
  • $\begingroup$ @Nex, What about part II? $\endgroup$ – ZFR Nov 22 '17 at 20:04
  • $\begingroup$ 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? $\endgroup$ – Nex Nov 22 '17 at 20:32
  • $\begingroup$ @ZFR I've made some edits to your post. Do you approve of this and agree to my amendments? $\endgroup$ – 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.

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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.

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

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