Preimage of Subgroup under Epimorphism is Subgroup
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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 epimorphism.
Let $H$ be a subgroup of $\struct {G_2, *}$.
Then:
- $\phi^{-1} \sqbrk H$ is a subgroup of $\struct {G_1, \circ}$
where $\phi^{-1} \sqbrk H$ denotes the preimage of $H$ under $\phi$.
Proof
Let $H$ be a subgroup of $\struct {G_2, *}$.
First note that from Null Relation is Mapping iff Domain is Empty Set:
- $\phi^{-1} \sqbrk H = \O \implies H = \O$
But $H \ne \O$.
Hence $\phi^{-1} \sqbrk H$ is not empty.
Next, let $x, y \in \phi^{-1} \sqbrk H$.
Then:
- $\exists h_1, h_2 \in H: h_1 = \map \phi x, h_2 = \map \phi y$
From the definition of Group Homomorphism, we have:
- $\map \phi {x^{-1} \circ y} = h_1^{-1} * h_2$
Since $H$ is a subgroup:
- $h_1^{-1} * h_2 \in H$
Hence:
- $x^{-1} \circ y \in \phi^{-1} \sqbrk H$
Thus from the One-Step Subgroup Test:
- $\phi^{-1} \sqbrk H$ is a subgroup of $G_1$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.20 \ \text {(b)}$