Group Epimorphism Preserves Normal Subgroups
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Theorem
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: G \to H$ be a group epimorphism.
Let $N \lhd G$, where $\lhd$ denotes that $N$ is a normal subgroup of $G$.
Then $\phi \sqbrk N \lhd H$.
That is, the image under $\phi$ of a normal subgroup is itself normal.
Proof
Let $N' := \phi \sqbrk N$.
From Group Homomorphism Preserves Subgroups, $N'$ is a subgroup of $H$.
It remains to show that $N'$ is normal in $H$.
Let $h \in H$ be arbitrary.
Let $n' \in N'$ be arbitrary.
Because $\phi$ is an epimorphism, it is by definition surjective.
Therefore:
- $\exists n \in N: \map \phi n = n'$
- $\exists g \in G: \map \phi g = h$
Then:
\(\ds g \circ n \circ g^{-1}\) | \(\in\) | \(\ds N\) | Definition of Normal Subgroup | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \phi {g \circ n \circ g^{-1} }\) | \(\in\) | \(\ds N'\) | as $N'$ is the image of $N$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \phi g * \map \phi n * \map \phi {g^{-1} }\) | \(\in\) | \(\ds N'\) | as $\phi$ has the morphism property | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \phi g * \map \phi n * \paren {\map \phi g}^{-1}\) | \(\in\) | \(\ds N'\) | Group Homomorphism Preserves Inverses | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds h * n' * h^{-1}\) | \(\in\) | \(\ds N'\) | Definition of $g$ and $n$ |
As $h$ and $n'$ were arbitrary, it follows that:
- $\forall h \in H, n' \in N': h * n' * h^{-1} \in N'$
and so $N'$ is a normal subgroup of $H$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.20 \ \text {(a)}$
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.10$: Theorem $29$