Internal Direct Products Isomorphic To External Direct Products

# Internal Direct Products Isomorphic To External Direct Products

Recall from The Internal Direct Product of Two Groups page that if $(G, *)$ is a group and $(H, *)$, $(K, *)$ are two subgroups of $(G, *)$ then we say that $(G, *)$ is the internal direct product of $(H, *)$ and $(K, *)$ if the following three conditions are met:

- $G = \{ h * k : h \in H, k \in K \}$.

- $H \cap K = \{ e \}$ where $e$ is the identity element in $G$.

- $h * k = k * h$ for all $h \in H$ and for all $k \in K$.

We will now look at a theorem which tells us that if $(G, *)$ is the internal direct product of $(H, *)$ and $(K, *)$ then $(G, *)$ is isomorphic to the external direct product $(H \times K, \cdot)$.

Theorem 1: Let $(G, *)$ be a group and let $(H, *)$, $(K, *)$ be subgroups of $(G, *)$. If $(G, *)$ is the internal direct product of $(H, *)$ and $(G, *)$ then $G \cong H \times K$. |

**Proof:**Define a function $f : H \times K \to G$ for all $(h, k) \in H \times K$ by:

\begin{align} \quad f(h, k) = h * k \end{align}

- We need show that $f$ is bijective. For injectivity, let $(h_1, k_1), (h_2, k_2) \in H \times K$ and suppose that $f(h_1, k_1) = f(h_2, k_2)$. Then:

\begin{align} \quad h_1 * k_1 = h_2 * k_2 \end{align}

- Thus:

\begin{align} \quad k_2^{-1} * k_1 = h_2 * h_1^{-1} \quad (*) \end{align}

- But $(H, *)$ and $(K, *)$ are groups and are closed under $*$ so $k_2^{-1} * k_1 \in K$ and $h_2 * h_1^{-1} \in H$. Since $H \cap K = \{ e \}$ this must mean that the equality at $(*)$ implies that $k_2^{-1} * k_1 = e$ and $h_2 * h_1^{-1} = e$. So $h_1 = h_2$ and $k_1 = k_2$ so $(h_1, k_1) = (h_2, k_2)$, i.e., $f$ is injective.

- We now show that $f$ is surjective. Let $g \in G$. Since $G$ is an interal direction product of $H$ and $K$ we have that $G = \{ h * k : h \in H, k \in K \}$ so $g = h * k$ for some $h \in H$ and for some $k \in K$. So $(h, k) \in H \times K$ is such that:

\begin{align} \quad f(h, k) = h * k = g \end{align}

- Hence $f$ is surjective.

- Lastly, we see that:

\begin{align} \quad f((h_1, k_1) \cdot (h_2, k_2)) = f(h_1 * h_2, k_1 * k_2) = (h_1 * h_2) * (k_1 * k_2) \end{align}

- Since $G$ is an internal direct product of $H$ and $K$ we see that $h_2 * k_1 = k_1 * h_2$ so:

\begin{align} \quad f((h_1, k_1) \cdot (h_2, k_2)) = (h_1 * k_1) * (h_2 * k_2) = f(h_1, k_1) * f(h_2, k_2) \end{align}

- Therefore $f$ is an isomorphism between $(G, *)$ and $(H \times K, \cdot)$ so $(G, *) \cong (H \times K, \cdot)$. $\blacksquare$