Isomorphism Preserves Semigroups/Proof 1
Jump to navigation
Jump to search
Theorem
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: S \to T$ be an isomorphism.
If $\struct {S, \circ}$ is a semigroup, then so is $\struct {T, *}$.
Proof
If $\struct {S, \circ}$ is a semigroup, then by definition it is closed.
From Morphism Property Preserves Closure, $\struct {T, *}$ is therefore also closed.
If $\struct {S, \circ}$ is a semigroup, then by definition $\circ$ is associative.
From Isomorphism Preserves Associativity, $*$ is therefore also associative.
So $\struct {T, *}$ is closed, and $*$ is associative, and therefore by definition, $\struct {T, *}$ is a semigroup.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups