Mapping to Square is Endomorphism iff Abelian

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Theorem

Let $\struct {G, \circ}$ be a group.

Let $\phi: G \to G$ be defined as:

$\forall g \in G: \map \phi g = g \circ g$


Then $\struct {G, \circ}$ is abelian if and only if $\phi$ is a (group) endomorphism.


Proof

Necessary Condition

Let $\struct {G, \circ}$ be an abelian group.

Let $a, b \in G$ be arbitrary.

Then:

\(\ds \map \phi {a \circ b}\) \(=\) \(\ds \paren {a \circ b} \circ \paren {a \circ b}\) Definition of $\phi$
\(\ds \) \(=\) \(\ds a \circ \paren {b \circ a} \circ b\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds a \circ \paren {a \circ b} \circ b\) Definition of Abelian Group: Commutativity
\(\ds \) \(=\) \(\ds \paren {a \circ a} \circ \paren {b \circ b}\) Group Axiom $\text G 1$: Associativity
\(\ds \) \(=\) \(\ds \map \phi a \circ \map \phi b\) Definition of $\phi$

As $a$ and $b$ are arbitrary, the above holds for all $a, b \in G$.

Thus $\phi$ is a group homomorphism from $G$ to $G$.

So by definition, $\phi$ is a group endomorphism.

$\Box$


Sufficient Condition

Let $\phi: G \to G$ as defined above be a group endomorphism.

Then:

\(\ds \forall a, b \in G: \, \) \(\ds \map \phi {a \circ b}\) \(=\) \(\ds \map \phi a \circ \map \phi b\) Definition of Group Endomorphism
\(\ds \leadsto \ \ \) \(\ds \forall a, b \in G: \, \) \(\ds \paren {a \circ b} \circ \paren {a \circ b}\) \(=\) \(\ds \paren {a \circ a} \circ \paren {b \circ b}\) Definition of $\phi$
\(\ds \leadsto \ \ \) \(\ds \forall a, b \in G: \, \) \(\ds a \circ \paren {b \circ a} \circ b\) \(=\) \(\ds a \circ \paren {a \circ b} \circ b\) Group Axiom $\text G 1$: Associativity
\(\ds \leadsto \ \ \) \(\ds \forall a, b \in G: \, \) \(\ds b \circ a\) \(=\) \(\ds a \circ b\) Cancellation Laws

Thus, by definition, $G$ is an abelian group.

$\blacksquare$


Sources

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