Assume
\(G,H\) are groups. When we defined the product
\(G \times H\text{,}\) we defined the surjective projection maps
\(\pi_G \colon G\times H \to G\) and
\(\pi_H \colon G \times H \to H\text{.}\) We show that the use of
\(\pi\) for projection here products coincides with the usage of
\(\pi\) for the projection to the quotient.
For every \(g \in G\text{,}\) we obtain an element \((g,1) \in G \times H\text{.}\) We observe there is a natural map
\begin{align*}
\mu_G \colon G \amp\to G \times H\\
x \amp \mapsto (x,1)
\end{align*}
that is a morphism of groups because
\begin{equation*}
\mu_G(x_1x_2) = (x_1x_2, 1) = (x_1,1)(x_2,1) = \mu_G(x_1)\mu_G(x_2)\text{.}
\end{equation*}
This morphism of groups is injective because
\begin{equation*}
\mu_G(x) = (x,1) = (1,1) \iff x = 1\text{.}
\end{equation*}
The image of \(G\) under \(\mu_G\) is the subgroup
\begin{equation*}
G \times 1 = \left\{(x,1) \;\middle\vert\; x \in G\right\} \leq G \times H\text{,}
\end{equation*}
and we note that \(G \cong G \times 1\text{.}\) This allows us to identify a copy of \(G\) inside \(G \times H\text{.}\) For all \((g,h) \in G \times H\text{,}\)
\begin{align*}
(g,h) \in \ker{\pi_H} \amp\iff \pi_H(g,h) = h = 1\\
\amp\iff h = 1\\
\amp\iff (g,h) \in G \times 1
\end{align*}
Since the projection \(\pi_H \colon G \times H \to H\) is surjective with kernel \(G \times 1\) and thus
\begin{equation*}
H \cong G\times H / G \times 1
\end{equation*}
by
Corollary 11.16. Note this implies that
\(G \times 1\) is normal in
\(G \times H\text{.}\)
