Group Homomorphism Preserves Inverses/Proof 1
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Theorem
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to\struct {H, *}$ be a group homomorphism.
Let:
Then:
- $\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$
Proof
Let $x \in G$.
Then:
\(\ds \map \phi x * \map \phi {x^{-1} }\) | \(=\) | \(\ds \map \phi {x \circ x^{-1} }\) | Definition of Group Homomorphism | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {e_G}\) | Definition of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds e_H\) | Group Homomorphism Preserves Identity |
So, by definition, $\map \phi {x^{-1} }$ is the right inverse of $\map \phi x$.
$\Box$
Similarly:
\(\ds \map \phi {x^{-1} } * \map \phi x\) | \(=\) | \(\ds \map \phi {x^{-1} \circ x}\) | Definition of Group Homomorphism | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {e_G}\) | Definition of Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds e_H\) | Group Homomorphism Preserves Identity |
So, again by definition, $\map \phi {x^{-1} }$ is the left inverse of $\map \phi x$.
$\Box$
Finally, as $\map \phi {x^{-1} }$ is both:
- a left inverse of $\map \phi x$
and:
- a right inverse of $\map \phi x$
it is by definition an inverse.
From Inverse in Group is Unique, $\map \phi {x^{-1} }$ is the only such element.
Hence the result.
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Morphisms
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 47.2$ Homomorphisms and their elementary properties
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms