Theorem 15.8

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Statement:

Let $G=H \times K$ be the direct product of groups $H$ and $K$. Then $\bar{H} = \{(h,e) | h \in H \}$ is a normal subgroup of $G$. Also $G/\bar{H}$ is isomorphic to $K$ in a natural way. Similarly, $G/\bar{K} \simeq H$ in a natural way.

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Proof:

Consider the homomorphism $\pi_2 : H \times K \rightarrow K$, where $\pi_2(h,k) = k$. Because $Ker(\pi_2)= \bar{H}$, we see that $\bar{H}$ is a normal subgroup of $H \times K$. Because $\pi_2$ is onto $K$, Theorem 14.11 tells us that $(H \times K)/\bar{H} \simeq K$.