# Multiplication by scalar axioms for an abelian group.

There is an R vector space where $k ⊙ x := x^k$ , $∀x, y ∈ V, k ∈ R$,

I showed that it was abelian. I wanted to show scalar multiplication by using the axioms.

1. $α ⊙ (x ⊕ y) = α ⊙ (xy) = (xy)^α = x^αy^α = (α ⊙ x)(α ⊙ y) = (α ⊙ x) ⊕ (α ⊙ y)$.

2. $(α + β) ⊙ x = x^{α+β} = x^αx^β = x^α ⊕ x^β = (α ⊙ x) ⊕ (β ⊙ x)$.

3. $(αβ) ⊙ x = x^{αβ} = (x^β)^α = α ⊙ x^β = α ⊙ (β ⊙ x)$

I don't understand what these 3 axioms show? distribution? or what exactly?Could someone explain to me what these 3 represent.

• Are you familiar with the ordinary axioms for a vector space, such as that of $\mathbb R^n$ over $\mathbb R$? Is it the concepts that you are stuck with or the notation? – Mathmo123 Aug 3 '14 at 20:14
• i am familiar its just that in this notation i feel like i dont even get wher they are showing distribution or whatever else they are showing.. @Mathmo123 – user135688 Aug 3 '14 at 20:21

I feel like the problem with this question is that the notation is obscuring things a little.

You have a vector space with two operations:

• Addition, which in this case is group multiplication - so $x \oplus y = x\cdot y$
• Scalar multiplication, which in this case satisfies $k \odot \alpha = \alpha^k$

I'll talk you through the first axiom. Let me know if you need help with the others.

The first axiom is distributivity over addition - in ordinary notation it would be $\alpha (x+y) = \alpha x + \alpha y$. Here, we prove this axiom holds

\begin{align}\color{blue} {\alpha \odot (x \oplus y) } &\color{blue}{= \alpha\odot(x\cdot y)} \qquad\text{ - as }x \oplus y = x\cdot y\\ &\color{blue} {=(x\cdot y)^\alpha}\qquad \text{ - by our definition of }k \odot \alpha = \alpha^k\\&\color{blue} {=x^\alpha \cdot y^\alpha}\qquad\text{- since the group is abelian}\\&\color{blue} {=(\alpha \odot x)\oplus(\alpha\odot y)}\qquad\text{ - as }x \oplus y = x\cdot y\end{align}

Notice that we specifically needed to use the fact that the group is abelian in this proof.

Similarly, $2$ is a proof of distributivity over multiplication. Can you interpret $3$?

• @Mathm0123 would it be associativity over multiplication? – user135688 Aug 4 '14 at 10:06
• Exactly. Does that make sense now? – Mathmo123 Aug 4 '14 at 13:28