Kernel of First Projection on Group Direct Product is Isomorphic to Second Factor

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Theorem

Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups.

Let $\struct {G, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$.

Let $\pr_1$ denote the first projection of $\struct {G, \circ}$:

$\forall \tuple {x, y} \in G: \map \phi {x, y} = x$


Then the kernel of $\pr_1$ is isomorphic to $\struct {G_2, \circ_2}$.


Proof

It is established in Projection on Group Direct Product is Epimorphism that $\pr_1$ is an epimorphism and so a homomorphism.

The kernel of $\pr_1$ is, by definition:

\(\ds \map \ker {\pr_1}\) \(=\) \(\ds \set {\tuple {x, y} \in G: \map \phi {x, y} = e_1}\) where $e_1$ is the identity of $\struct {G_1, \circ_1}$
\(\ds \) \(=\) \(\ds \set {\tuple {x, y} \in G: x = e_1}\) Definition of $\pr_1$
\(\ds \) \(=\) \(\ds \set {\tuple {e_1, y}: y \in G_2}\) Definition of $\pr_1$

Let $H := \set {\tuple {e_1, y}: y \in G_2}$.

It is to be established that $\struct {H, \circ}$ is isomorphic to $\struct {G_2, \circ_2}$.

Let $\theta: H \to G_2$ be defined as:

$\forall \tuple {e_1, y} \in H: \map \theta {e_1, y} = y$

We have that $\theta$ is the second projection on $H$, and so itself an epimorphism, and so surjective.

Then:

\(\ds \map \theta {e_1, y_a}\) \(=\) \(\ds \map \theta {e_1, y_b}\) for $y_a, y_b \in G_2$
\(\ds \leadsto \ \ \) \(\ds y_a\) \(=\) \(\ds y_b\) Definition of $\theta$
\(\ds \leadsto \ \ \) \(\ds \tuple {e_1, y_a}\) \(=\) \(\ds \tuple {e_1, y_b}\)

demonstrating that $\theta$ is an injection.

So $\theta$ is a homomorphism which is both surjective and injective.

Therefore $\theta$ is an isomorphism from $\map \ker {\pr_1}$ to $\struct {G_2, \circ_2}$.

Hence the result.

$\blacksquare$


Sources

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