# Subset of Module Containing Identity is Linearly Dependent

## Theorem

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

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.

Let $\struct {G, +_G, \circ}_R$ be an $R$-module.

Let $H \subseteq G$ such that $e \in H$.

Then $H$ is a linearly dependent set.

## Proof

From Scalar Product with Identity, $\forall \lambda: \lambda \circ e = e$.

Let $H \subseteq G$ such that $e \in H$.

Consider any sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ in $H$ which includes $e$.

So, let $a_j = e$ for some $j \in \closedint 1 n$.

Let $c \in R \ne 0_R$.

Consider the sequence $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ of elements of $R$ defined as:

$\lambda_k = \begin{cases} c & : k \ne j \\ 0_R & : k= j \end{cases}$

Then:

 $\ds \sum_{k \mathop = 1}^n \lambda_k \circ a_k$ $=$ $\ds \lambda_1 \circ a_1 + \lambda_2 \circ a_2 + \cdots + \lambda_j \circ a_j + \cdots + \lambda_n \circ a_n$ $\ds$ $=$ $\ds 0_R \circ a_1 + 0_R \circ a_2 + \cdots + c \circ e + \cdots + 0_R \circ a_n$ $\ds$ $=$ $\ds e + e + \cdots + e + \cdots + e$ $\ds$ $=$ $\ds e$

Thus there exists a sequence $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ in which not all $\lambda_k = 0_R$ such that:

$\ds \sum_{k \mathop = 1}^n \lambda_k \circ a_k = e$

Hence the result.

$\blacksquare$