# Semigroup Isomorphism/Examples/Structure with Two Operations

## Examples of Semigroup Isomorphisms

Let $\struct {S_1, \circ_1, *_1}$ and $\struct {S_2, \circ_2, *_2}$ be algebraic structures such that:

$\struct {S_1, \circ_1}$ is isomorphic to $\struct {S_2, \circ_2}$
$\struct {S_1, *_1}$ is isomorphic to $\struct {S_2, *_2}$

Then it is not necessarily the case that $\struct {S_1, \circ_1, *_1}$ is isomorphic to $\struct {S_2, \circ_2, *_2}$.

## Proof

Proof by Counterexample

Let $\R$ denote the set of real numbers.

Let $\vee$ and $\wedge$ denote the max operation and min operation respectively.

Let $\struct {\R, \vee}$ and $\struct {\R, \wedge}$ denote the algebraic structures formed from the above.

From Max and Min Operations on Real Numbers are Isomorphic, $\struct {\R, \vee}$ is isomorphic to $\struct {\R, \wedge}$.

Let $\struct {\R, \times}$ denote the algebraic structure formed from $\R$ under multiplication .

From Real Numbers under Multiplication form Monoid, $\struct {\R, \times}$ is a monoid and hence a semigroup.

Let $I_\R: \R \to \R$ denote the identity mapping on $\R$.

From Identity Mapping is Semigroup Automorphism we have that $\struct {\R, \times}$ is an automorphism and hence a fortiori an isomorphism.

Now consider the algebraic structures $\struct {\R, \vee, \times}$ and $\struct {\R, \wedge, \times}$.

We have from above that

We also have that:

$\struct {\R, \vee}$ is isomorphic to $\struct {\R, \wedge}$
$\struct {\R, \times}$ is isomorphic to $\struct {\R, \times}$

Aiming for a contradiction, suppose there exists an isomorphism $\phi$ from $\struct {\R, \vee, \times}$ to $\struct {\R, \wedge, \times}$.

Because $\phi$ is an isomorphism, it is by definition a bijection

Hence $\phi$ is both a surjection and an injection.

Then:

 $\ds \exists x \in \R: \,$ $\ds \map \phi x$ $=$ $\ds 1$ Definition of Surjection $\ds \map \phi 0$ $=$ $\ds \map \phi {x \times 0}$ $\ds$ $=$ $\ds \map \phi x \times \map \phi 0$ Definition of Semigroup Isomorphism $\ds$ $=$ $\ds 1 \times \map \phi 0$ $\ds$ $=$ $\ds 0$

Then:

 $\ds \map \phi 1$ $=$ $\ds \map \phi {1 \vee 0}$ Definition of $\vee$ $\ds$ $=$ $\ds \map \phi 1 \wedge \map \phi 0$ Definition of Isomorphism (Abstract Algebra) $\ds$ $=$ $\ds 0$ Definition of $\wedge$ and from $\map \phi 0 = 0$ from above

Thus we have that:

$\map \phi 0 = 0$

and:

$\map \phi 1 = 0$

and so $\phi$ is not an injection.

This contradicts our assertion that isomorphism.

Hence by Proof by Contradiction no such isomorphism exists.

Hence $\struct {\R, \vee, \times}$ and $\struct {\R, \wedge, \times}$ are not isomorphic.

The result follows.

$\blacksquare$