# Equivalence of Definitions of Independent Subgroups

## Theorem

The following definitions of the concept of Independent Subgroups are equivalent:

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

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

### Definition 1

The subgroups $H_1, H_2, \ldots, H_n$ are independent if and only if:

$\displaystyle \prod_{k \mathop = 1}^n h_k = e \iff \forall k \in \left\{{1, 2, \ldots, n}\right\}: h_k = e$

where $h_k \in H_k$ for all $k \in \left\{{1, 2, \ldots, n}\right\}$.

### Definition 2

The subgroups $H_1, H_2, \ldots, H_n$ are independent if and only if:

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

## Proof

### Definition 1 implies Definition 2

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

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

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

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

Then:

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

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 $j \in \left\{{k+1, k+2, \ldots, n}\right\}$ let $x_j = e$.

Then:

 $\displaystyle \prod_{j \mathop = 1}^n x_j$ $=$ $\displaystyle \left({\prod_{j \mathop = 1}^{k-1} x_j}\right) u^{-1} \left({\prod_{j \mathop = k+1}^{n} x_j}\right)$ $\displaystyle$ $=$ $\displaystyle \left({\prod_{j \mathop = 1}^{k-1} x_j}\right) u^{-1}$ $\displaystyle$ $=$ $\displaystyle e$

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

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

$\displaystyle \forall k \in \left\{{2, 3, \ldots, n}\right\}: \left({\prod_{j \mathop = 1}^{k-1} H_j}\right) \cap H_k = \left\{{e}\right\}$

### Definition 2 implies Definition 1

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

$\displaystyle \forall k \in \left\{{2, 3, \ldots, n}\right\}: \left({\prod_{j \mathop = 1}^{k-1} H_j}\right) \cap H_k = \left\{{e}\right\}$

Aiming for a contradiction, suppose that:

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

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

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:

 $\displaystyle e$ $=$ $\displaystyle x_1 \cdots x_n$ $\displaystyle$ $=$ $\displaystyle x_1 \cdots x_m$ $\displaystyle$ $=$ $\displaystyle \left({x_1 \cdots x_{m-1} }\right) x_m$

Thus:

 $\displaystyle x_m^{-1}$ $=$ $\displaystyle x_1 \cdots x_{m-1}$ $\displaystyle$ $=$ $\displaystyle H_1 \cdots H_{m-1}$

So:

$x_m^{-1} \in H_m$

Therefore by hypothesis:

$x_m^{-1} = e$

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:

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

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

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

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

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

$\blacksquare$