3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.