Zero Vector Space Product iff Factor is Zero/Proof 1

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $\left({K, +, \circ}\right)$ be a division ring whose zero is $0$ and whose unity is $1$.

Let $\left({G, +_G, \circ}\right)_K$ be a $K$-vector space.

Let $x \in G, \lambda \in K$.

Then $\lambda \circ x = e \iff \left({\lambda = 0 \lor x = e}\right)$.

Proof
A vector space is a module, so all results about modules also apply to vector spaces.

So from Scalar Product with Identity it follows directly that $\lambda = 0 \lor x = e \implies \lambda \circ x = e$.

Next, suppose $\lambda \circ x = e$ but $\lambda \ne 0$.

Then from Scalar Product with Identity: