Quotient Group of Direct Products

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Theorem

Let $G$ and $G'$ be groups.

Let:

$H \triangleleft G$
$H' \triangleleft G'$

where $\triangleleft$ denotes the relation of being a normal subgroup.

Then:

$(1): \quad \left({H \times H'}\right) \triangleleft \left({G \times G'}\right)$
$(2): \quad \left({G \times G'}\right) / \left({H \times H'}\right)$ is isomorphic to $\left({G / H}\right) \times \left({G' / H'}\right)$

where:


Proof

$(1): \quad \left({H \times H'}\right) \triangleleft \left({G \times G'}\right)$:

Let $\left({x, x'}\right) \in G \times G'$ and $\left({y, y'}\right) \in H \times H'$.

Then:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \left({x, x'}\right) \left({y, y'}\right) \left({x, x'}\right)^{-1}\) \(=\) \(\displaystyle \left({x, x'}\right) \left({y, y'}\right) \left({x^{-1}, x'^{-1} }\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left({x y x^{-1}, x' y' x'^{-1} }\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\in\) \(\displaystyle H \times H'\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          by the normality of $H$ and $H'$.          


$(2): \quad \left({G \times G'}\right) / \left({H \times H'}\right)$ is isomorphic to $\left({G / H}\right) \times \left({G' / H'}\right)$:

Let $\varphi_1 : G \to G / H$ and $\varphi_2 : G' \to G' / H'$ be the quotient epimorphisms with $H$ and $H'$ as their kernels, respectively.

Now define a homomorphism $\varphi : G \times G' \to \left({G / H}\right) \times \left({G' / H'}\right)$ by $\varphi = \varphi_1 \times \varphi_2$.

The kernel of $\varphi$ is clearly $H \times H'$, and $\varphi$ is surjective.

So $\left({G / H}\right) \times \left({G' / H'}\right) \cong \left({G \times G'}\right) / \left({H \times H'}\right)$.



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