Elements of Group with Equal Images under Homomorphisms form Subgroup
Jump to navigation
Jump to search
Theorem
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $f: G \to H$ and $g: G \to H$ be group homomorphisms.
Then the set:
- $S = \set {x \in G: \map f x = \map g x}$
is a subgroup of $G$.
Proof
Let the identities of $\struct {G, \circ}$ and $\struct {H, *}$ be $e_G$ and $e_H$ respectively.
By Homomorphism to Group Preserves Identity:
- $\map f {e_G} = \map g {e_G} = e_H$
Thus $e_G \in S$, and so $S \ne \O$.
Similarly, from Homomorphism to Group Preserves Inverses, $x \in S \implies x^{-1} \in S$.
Let $x, y \in S$. Then:
\(\ds \map f {x \circ y}\) | \(=\) | \(\ds \map f x * \map f y\) | Morphism Property | |||||||||||
\(\ds \) | \(=\) | \(\ds \map g x * \map g y\) | Definition of $S$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map g {x \circ y}\) | Morphism Property |
Thus $x \circ y \in S$.
So, by the Two-Step Subgroup Test:
- $S \le G$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.8$