Definition:Independent Subgroups

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

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

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

The subgroups $$H_1, H_2, \ldots, H_n$$ are described as independent iff:

$$\prod_{k=1}^n h_k = e \iff \forall k \in \left[{1 \,. \, . \, n}\right]: h_k = e$$

That is, the product of any elements from different $$H_k$$ instances forms the identity only if all of those elements are the identity.

Then $$H_1, H_2, \ldots, H_n$$ are independent iff $$\forall k \in \left[{2 \,. \, . \, n}\right]: \left({\prod_{j=1}^{k-1} H_j}\right) \cap H_k = \left\{{e}\right\}$$.

Proof

 * Necessity:

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

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

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 $$u = e$$.


 * Sufficiency:

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

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

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.