Complement of Normal Subgroup is Isomorphic to Quotient Group
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Definition
Let $G$ be a group with identity $e$.
Let $N$ be a normal subgroup of $G$.
Let $K$ be a complement of $N$.
Then the quotient group $G / N$ is isomorphic to $K$.
Proof
Let $\phi: K \to G / N$ be the mapping defined as:
- $\forall k \in K: \map \phi k = k N$
We have:
\(\ds \forall x, y \in K: \, \) | \(\ds \map \phi {x y}\) | \(=\) | \(\ds \paren {x y} N\) | Definition of $\phi$ | ||||||||||
\(\ds \) | \(=\) | \(\ds x N y N\) | Definition of Coset Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi x \, \map \phi y\) | Definition of $\phi$ |
Thus $\phi$ is shown to be a homomorphism.
By definition of $K$ and $N$ we have that:
- $\forall g \in G: \exists k \in K, n \in N: g = k n$
That is, every element of $g$ is the product of an element of $K$ with an element of $N$.
Let $x N, y N \in G / N$.
Then:
\(\ds x N\) | \(=\) | \(\ds y N\) | Definition of $\phi$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds k_1 n_1 N\) | \(=\) | \(\ds k_2 n_2 N\) | for some $k_1, k_2 \in K$, $n_1 n_2 \in N$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds k_1 N\) | \(=\) | \(\ds k_2 N\) | as $n_1 N = N$ and $n_2 N = N$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \phi {k_1}\) | \(=\) | \(\ds \map \phi {k_2}\) | Definition of $\phi$ |
demonstrating that $\phi$ is an injection.
Then we have:
\(\ds x N\) | \(\in\) | \(\ds G / N\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds k n N\) | \(\in\) | \(\ds G / N\) | for some $k \in K$, $n \in N$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds k N\) | \(\in\) | \(\ds G / N\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists k \in K: \, \) | \(\ds k N\) | \(=\) | \(\ds \map \phi k\) | Definition of $\phi$ |
demonstrating that $\phi$ is a surjection.
Thus $\phi$ is an injective and surjective homomorphism.
Hence, by definition, $\phi$ is an isomorphism.
Hence the result.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $22$