Equivalence of Definitions of Independent Subgroups

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

Let $\sequence {H_n}$ be a sequence of independent subgroups of $G$.

Definition 1 implies Definition 2
Let $H_1, H_2, \ldots, H_n$ be independent by definition 1:


 * $\ds \prod_{k \mathop = 1}^n h_k = e \iff \forall k \in \closedint 1 n: h_k = e$

where $h_k \in H_k$ for all $k \in \closedint 1 n$.

Let $\ds u \in \paren {\prod_{j \mathop = 1}^{k - 1} H_j} \cap H_k$.

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

We have that $u \in H_k$ and $H_k$ is a group.

Therefore let $x_k = u^{-1}$, the inverse of $u$.

For each $j \in \set {k + 1, k + 2, \ldots, n}$ let $x_j = e$.

Then:

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

Hence $H_1, H_2, \ldots, H_n$ are independent by definition 2:


 * $\ds \forall k \in \set {2, 3, \ldots, n}: \paren {\prod_{j \mathop = 1}^{k - 1} H_j} \cap H_k = \set e$

Definition 2 implies Definition 1
Let $H_1, H_2, \ldots, H_n$ be independent by definition 2:


 * $\ds \forall k \in \set {2, 3, \ldots, n}: \paren {\prod_{j \mathop = 1}^{k - 1} H_j} \cap H_k = \set e$

that:


 * $\ds x_1, x_2, \ldots, x_n \in H_1, H_2, \ldots, H_n: \prod_{i \mathop = 1}^n x_i = e$

but that $x_j \ne e$ for some $j \in \set {1, 2, \ldots, n}$.

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

So let $m$ be the largest of the $j$ such that $x_j \ne e$.

Then $m > 1$, and:

Thus:

So:
 * $x_m^{-1} \in H_m$

Therefore :
 * $x_m^{-1} = e$

and therefore:
 * $x_m = e$

But $m$ was defined to be the largest of the $j$ such that $x_j \ne e$.

Thus by Proof by Contradiction it follows that there can be no such $j$ so that:
 * $\ds x_1, x_2, \ldots, x_n \in H_1, H_2, \ldots, H_n: \prod_{i \mathop = 1}^n x_i = e$

but that $x_j \ne e$ for some $j \in \set {1, 2, \ldots, n}$.

Hence $H_1, H_2, \ldots, H_n$ are independent by definition 1:


 * $\ds \prod_{k \mathop = 1}^n h_k = e \iff \forall k \in \closedint 1 n: h_k = e$

where $h_k \in H_k$ for all $k \in \closedint 1 n$.