# Condition on Congruence Relations for Cancellable Monoid to be Group/Counterexample

## Theorem

Let $\struct {S, \circ}$ be a monoid which is not cancellable.

Let every non-trivial congruence relation on $\struct {S, \circ}$ be induced by a normal subgroup of $\struct {S, \circ}$.

Then it is not necessarily the case that $\struct {S, \circ}$ is a group.

## Proof

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Consider the Multiplicative Monoid of Integers Modulo $3$ $\struct {\Z_3, \times_3}$, defined by its Cayley table:

- $\begin{array} {r|rrr} \struct {\Z_3, \times_3} & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 & \\ \hline \eqclass 0 3 & \eqclass 0 3 & \eqclass 0 3 & \eqclass 0 3 \\ \eqclass 1 3 & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\ \eqclass 2 3 & \eqclass 0 3 & \eqclass 2 3 & \eqclass 1 3 \\ \end{array}$

which can also be presented:

- $\begin{array} {r|rrrrr} \times_3 & 0 & 1 & 2 \\ \hline 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 \\ 2 & 0 & 2 & 1 \end{array}$

It is noted that $\struct {\Z_3, \times_3}$ is not a cancellable monoid.

Indeed:

- $0 \times_3 1 = 0$
- $0 \times_3 2 = 0$

and so $\times_3$ is not cancellable for $0$.

Hence from Group is Cancellable Monoid:

- $\struct {\Z_3, \times_3}$ is not a group.

We note that $\struct {\Z_3, \times_3}$ has an identity element $\eqclass 1 3$.

We list all the equivalence relations on $\struct {\Z_3, \times_3}$ by the partitions they induce:

- $\set {\set 0, \set 1, \set 2}$ is induced by the diagonal relation
- $\set {\set {0, 1, 2}}$ is induced by the trivial relation
- $\set {\set {0, 1}, \set 2}$ is not a congruence relation because we have $0 \mathrel \RR 1$ and $2 \mathrel \RR 2$ but not $\paren {0 \times 2} \mathrel \RR \paren {1 \times 2}$
- $\set {\set {0, 2}, \set 1}$ is not a congruence relation because we have $0 \mathrel \RR 2$ and $2 \mathrel \RR 2$ but not $\paren {0 \times 2} \mathrel \RR \paren {2 \times 2}$

- $\set {\set 0, \set {1, 2}}$ is induced by the equivalence relation induced by the normal subgroup $\set 0$.

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We see that the first four relations are not non-trivial congruence relation on $\struct {S, \circ}$.

Therefore every non-trivial congruence relation on $\struct {S, \circ}$ is induced by a normal subgroup of $\struct {S, \circ}$.

But, *a priori*, $\struct {\Z_3, \times_3}$ is not a group.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.18 \ \text {(b)}$