Subset of Module Containing Identity is Linearly Dependent

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

Let $\left({R, +, \circ}\right)$ be a ring whose zero is $0_R$.

Let $\left({G, +_G, \circ}\right)_R$ 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 $\left \langle {a_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ in $H$ which includes $e$.

So, let $a_j = e$ for some $j \in \left[{1 \,.\,.\, n}\right]$.

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

Consider the sequence $\left \langle {\lambda_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ of elements of $R$ defined as:
 * $\lambda_k = \begin{cases}

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

Then:

Thus there exists a sequence $\left \langle {\lambda_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ in which not all $\lambda_k = 0_R$ such that $\displaystyle \sum_{k \mathop = 1}^n \lambda_k \circ a_k = e$,.

Hence the result.