Group Homomorphism Preserves Inverses/Proof 3

Theorem

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.

Let $\phi: \struct {G, \circ} \to\struct {H, *}$ be a group homomorphism.

Let:

$e_G$ be the identity of $G$

$e_H$ be the identity of $H$

Then:

$\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$

Proof

From Group Homomorphism of Product with Inverse, we have:

$\forall x, y \in G: \map \phi {x \circ y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$

Putting $x = e_G$ and $y = x$ we have:

\(\ds \map \phi {x^{-1} }\)

\(=\)

\(\ds \map \phi {e_G \circ x^{-1} }\)

\(\ds \)

\(=\)

\(\ds \map \phi {e_G} * \paren {\map \phi x}^{-1}\)

\(\ds \)

\(=\)

\(\ds e_H * \paren {\map \phi x}^{-1}\)

Group Homomorphism Preserves Identity

\(\ds \)

\(=\)

\(\ds \paren {\map \phi x}^{-1}\)

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Corollary $8.7 \ \text{(ii)}$