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Here is Prob. 14, Sec. 2.3, in the book Topics in Algebra by I.N. Herstein, 2nd edition:

Suppose a finite set $G$ is closed under an associative product and that both cancellation laws hold in $G$. Prove that $G$ must be a group.

Here is a Mathematics Stack Exchange post on this problem.

And, here is Prob. 16, Sec. 2.3, in the book Topics in Algebra by I.N. Herstein, 2nd edition:

In Problem 14 show by an example that if one just assumed one of the cancellation laws, then the conclusion need not hold.

My Attempt:

Let $$ G = \{ \, 1, -1 \, \}. $$ And for any $a, b \in G$, let $a * b$ be defined as $$ a*b \colon= a \, \lvert b \rvert. $$ Thus we have $$ 1 * 1 = 1 = 1 * (-1), \qquad (-1)*1 = -1 = (-1) * (-1). $$

We note that in this set $G$ with this binary operation $*$, the left cancellation law does not hold, for although we have $$ 1 * 1 = 1 = 1 * (-1), $$ it is not true that $$ 1 = -1.$$

However, the right cancellation law does indeed hold in $G$, for if $a$, $b$, and $c$ are any elements in $G$ and if $$ a * c = b * c, $$ then we have $$ a \, \lvert c \rvert = b \, \lvert c \rvert, $$ and since $c \neq 0$, therefore we must have $$ a = b. $$

Thus in our set $G$ with this particular binary operation $*$, the right cancellation law holds but the left cancellation law does not.

Moreover, for any elements $a$, $b$, and $c$ in $G$, we obtain $$ \begin{align} (a * b) * c &= \big(a \, \lvert b \rvert \big) * c \\ &= \big(a \, \lvert b \rvert \big) \, \lvert c \rvert \\ &= a \, \lvert b c \rvert \\ &= a \, \big\lvert b \, \lvert c \rvert \, \big\rvert \\ &= a \, \big\lvert \, b * c \, \big\rvert \\ &= a * ( b * c). \end{align} $$ This shows that $*$ is an associative binary operation on $G$.

Thus $G$ is a finite set with an associative binary operation $*$ such that only one of the two cancellation laws holds in $G$.

Finally, we note that this set $G$ with this particular binary operation $*$ is not a group, because $G$ contains no identity element.

In order to see that $G$ with respect to $*$ contains no identity element, we note that as $$ 1 * (-1) \neq -1, $$ so $1$ cannot be an identity element (more precisely a left identity element) of $G$, and as $$ (-1) * 1 \neq 1, $$ so $-1$ cannot be an identity element (more precisely, a left identity element) of $G$.

On the other hand, either of $1$ and $-1$ can serve as a right identity element in $G$, because for any element $a \in G$, we have $$ a * (\pm 1) = a \, \lvert \pm 1 \rvert = a (1) = a. $$

However, in a group a right identity element is also a left identity element, and there is a unique such element. See Lemma 2.3.1 a. in Herstein. And, also refer to this Mathematics Stack Exchange post.

Is this example good enough? If so, then is my presentation clear enough too? Or, are there any issues?

Now let us consider this same set $G$, namely, $$ G = \{ \, 1, -1 \, \}, $$ but now let $*$ be the binary operation on $G$ defined as follows: For any elements $a$ and $b$ of $G$, let us define $a ~ b$ as $$ a*b \colon= \lvert a \rvert \, b. $$ Then we have $$ 1 * 1 = 1 = (-1) * 1, \qquad 1 * (-1) = -1 = (-1) * (-1). $$ In this case, the left cancellation law holds in $G$, for if $a, b, c \in G$ and if $$ a * b = a * c, $$ then $$ \lvert a \rvert \, b = \lvert a \rvert \, c, $$ and as $a \neq 0$, so we must have $b = c$, but the right does not hold in $G$, for we have $$ 1 * 1 = (-1) * 1, $$ but $1 \neq -1$. And, this binary operation $*$ on $G$ is of course associative, for if $a, b, c \in G$, then we obtain $$ \begin{align} a * ( b * c ) &= a * \big( \lvert b \rvert \, c \big) \\ &= \lvert a \rvert \, \big( \lvert b \rvert \, c \big) \\ &= \lvert a \rvert \, \lvert b \rvert \, c \\ &= \lvert a b \rvert \, c \\ &= \big\lvert a \, \lvert b \rvert \big\rvert \, c \\ &= \big\lvert a * b \big\rvert \, c \\ &= (a * b) *c. \end{align} $$

Thus again we have the finite set $G$ with an associative binary operation such that only one of the two cancellation laws holds in $G$.

Finally, we note that this set $G$ with this particular binary operation $*$ is again not a group. The element $1$ of $G$ cannot serve as an identity element (from the right) because $(-1) * 1 \neq -1$, and $-1$ cannot serve as an identity element (again from the right) because $1 * (-1) \neq 1$, although either of $1$ and $-1$ can serve as an identity element from the left, because for any element $a \in G$ have $$ (\pm 1 ) * a = \lvert \pm 1 \rvert \, a = 1 \cdot a = a. $$

Is this example OK? Or, are there problems?

Any other examples of this kind please?

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  • $\begingroup$ You have a typo: you say $b=c$ when you actually mean $a=b$. But I think the proof is otherwise both correct and well written. $\endgroup$ – Robert Shore Oct 27 '19 at 9:33
  • $\begingroup$ @RobertShore I've corrected that typo. Thank you for reviewing my post. I've just seen your profile. I too want to pursue a law degree, or at least learn the core law subjects. $\endgroup$ – Saaqib Mahmood Oct 27 '19 at 11:29
  • $\begingroup$ This is correct. $\endgroup$ – Matt Samuel Oct 27 '19 at 15:18

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