Group Homomorphism Preserves Subgroups
Jump to navigation
Jump to search
Theorem
Let $\struct {G_1, \circ}$ and $\struct {G_2, *}$ be groups.
Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ be a group homomorphism.
Then:
- $H \le G_1 \implies \phi \sqbrk H \le G_2$
where:
- $\phi \sqbrk H$ denotes the image of $H$ under $\phi$
- $\le$ denotes subgroup.
That is, group homomorphism preserves subgroups.
Proof
Let $H \le G_1$.
First note that from Null Relation is Mapping iff Domain is Empty Set:
- $H \ne \O \implies \phi \sqbrk H \ne \O$
and so $\phi \sqbrk H$ is not empty.
Next, let $x, y \in \phi \sqbrk H$.
Then:
- $\exists h_1, h_2 \in H: x = \map \phi {h_1}, y = \map \phi {h_2}$
From the definition of Group Homomorphism, we have:
- $\map \phi {h_1^{-1} \circ h_2} = x^{-1} * y$
Since $H$ is a subgroup:
- $h_1^{-1} \circ h_2 \in H$
Hence:
- $x^{-1} * y \in \phi \sqbrk H$
Thus from the One-Step Subgroup Test:
- $\phi \sqbrk H \le G_2$
$\blacksquare$
Sources
- 1974: Thomas W. Hungerford: Algebra ... (previous) ... (next): $\S 1.2$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 47.3$ Homomorphisms and their elementary properties