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.

$\blacksquare$


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