How to get $(ab)1 = (a1)(b1)$ in Galois field?

Question

I'm reading Galois field from textbook Groups, Matrices, and Vector Spaces - A Group Theoretic Approach to Linear Algebra by James B. Carrell.

Here $r,a,b \in \mathbb N$ and $1 \in \mathbb F$.

While I understand $(ab)1 = a(b1)$, I could not get $(ab)1 = (a1)(b1)$. Clearly, $(ab)1$ means we sum $1 \in \mathbb F$ repeatedly $ab$ times, whereas $(a1)(b1)$ means the multiplication of $a1 \in \mathbb F$ and $b1 \in\mathbb F$.

Could you please elaborate on this equality?

Answer 1

By the distributive property, we have $$ (a1)(b1) = (\overbrace{1 + \cdots + 1}^{a \text{ times}})(b1) = \overbrace{b1 + \cdots + b1}^{a \text{ times}} = a(b1) = (ab)1. $$