Equivalence of Definitions of Independent Subgroups

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

Let $\left \langle {H_n} \right \rangle$ be a sequence of independent subgroups of $G$.

Then:
 * $H_1, H_2, \ldots, H_n$ are independent

iff:
 * $\displaystyle \forall k \in \left[{2 \,.\,.\, n}\right]: \left({\prod_{j \mathop = 1}^{k-1} H_j}\right) \cap H_k = \left\{{e}\right\}$

where $\left[{m \,.\,.\, n}\right]$ is to be interpreted as the (closed) integer interval from $m$ to $n$.

Proof
By definition, the subgroups $H_1, H_2, \ldots, H_n$ are described as independent iff:
 * $\displaystyle \prod_{k \mathop = 1}^n h_k = e \iff \forall k \in \left[{1 \,.\,.\, n}\right]: h_k = e$

where $h_k \in H_k$ for all $k \in \left[{1 \,.\,.\, n}\right]$.

Necessary Condition
Let $\displaystyle u \in \left({\prod_{j \mathop = 1}^{k-1} H_j}\right) \cap H_k$.

Then $\displaystyle \exists x_1, x_2, \ldots, x_{k-1} \in H_1, \ldots, H_{k-1}: u = \prod_{j \mathop = 1}^{k-1} x_j$.

Let $x_k = u^{-1}$, which is justified because $u \in H_k$ and $H_k$ is a group.

Also, let $x_j = e$ for each $j \in \left[{k+1 \,.\,.\, n}\right]$. Then:

Thus by hypothesis $u^{-1} = e$ and hence $u = e$.

Sufficient Condition
Suppose $\displaystyle x_1, x_2, \ldots, x_n \in H_1, H_2, \ldots, H_n: \prod_{i \mathop = 1}^n x_i = e$ but that $x_j \ne e$ for some $j \in \left[{1 \,.\,.\, n}\right]$.

There is bound to be more than one of them, because otherwise that would mean $\displaystyle \prod_{i \mathop = 1}^n x_i = x_j e = e \implies x_j = e$.

So let $m$ be the largest of the $j$ such that $x_j \ne e$. Then $m > 1$, and:

Thus:

So $x_m^{-1} \in H_m$, so $x_m^{-1} = e$ by hypothesis, and therefore $x_m = e$.

The result follows.