# Internal Direct Product Theorem/General Result/Proof 2

## Theorem

Let $G$ be a group whose identity is $e$.

Let $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of subgroups of $G$.

Then $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ if and only if:

$(1): \quad G = H_1 H_2 \cdots H_n$
$(2): \quad \sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups
$(3): \quad \forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$

where $H_k \lhd G$ denotes that $H_k$ is a normal subgroup of $G$.

## Proof

It is to be shown that:

$(1): \quad$ Each $H_1, H_2, \ldots, H_n$ is a normal subgroup of $G$
$(2): \quad$ Each element $g$ of $G$ can be expressed uniquely in the form:
$g = h_1 \circ h_2 \circ \cdots \circ h_n$
where $h_i \in H_i$ for all $i \in \set {1, 2, \ldots, n}$.

Condition $(3)$ already gives that $H_i$ is normal for all $i \in \set {1, 2, \ldots, n}$.

Condition $(1)$ gives us that each element $g$ of $G$ can be expressed in the form:

$g = h_1 h_2 \dotsm h_n$ with $h_i \in H_i$ for all $i \in \set {1, 2, \ldots, n}$.

It is now shown that this expression is unique.

Suppose that:

$g = h_1 h_2 \dotsm h_n = k_1 k_2 \dotsm k_n$

where $h_i, k_i \in H_i$ for all $i \in \set {1, 2, \ldots, n}$ and $h_j \ne k_j$ for at least one $j$.

Let $j$ be the largest integer such that $h_j \ne k_j$, so that $h_i = k_i$ for $i > j$.

Cancelling $h_i$ for $i > j$ gives:

$h_i h_2 \dotsm h_j = k_1 k_2 \dotsm k_j$

and so:

$h_j {k_j}^{-1} = \paren {h_1 h_2 \dotsm h_{j - 1} }^{-1} \paren {k_1 k_2 \dotsm k_{j - 1} } \in \paren {H_1 H_2 \dotsm H_{j - 1} } \cap G_j$

But by condition $(2)$:

$\paren {H_1 H_2 \dotsm H_{j - 1} } \cap G_j = \set e$

by definition of independent subgroups.

Thus $h_j = k_j$, which contradicts our assertion that $h_j \ne k_j$.

Hence the decomposition is unique.

$\blacksquare$