Groups: n in Additive Notation

Question

My textbook states,

In additive notation, we denote $a+a+a+ \dots +a$ for $n$ summands by $na$.

That makes sense to me. However, it then notes,

Be careful: In the notation $na$, the number $n$ is in $\mathbb{Z}$, not in $G$. One reason we prefer to present group theory using multiplicative notation, even if $G$ is abelian, is the confusion caused by regarding $n$ as being in $G$ in this notation $na$. No one ever misinterprets the $n$ when it appears in an exponent.

What exactly does this mean? I don’t understand it. Why must $n$ be in $\mathbb{Z}$ rather than in $G$? Thank you.

Answer 1

The group $G$ might be $V_4$, for example, the four group of Klein. Thus $G=\{e,a,b,c\}$ so you can look at $a\cdot a\cdot a=3 a$ using the first notation, but then you might think $3\in V_4$, which is definitely not the case. Thus, for the sake of clarity, we write $a^3$. In a group, you have only one operation which we typically denote by $+, \circ, \cdot, \square$ et cetera, so if you write $a\cdot a\cdot a$ as $a^3$, you know unambiguously what is meant.

Answer 2

$a+a=2a$

$a+a+a=3a$

$a+a+a+a=4a$

$\underbrace{a+a+\ldots+a}_{n \text{ times }}=na$

$\underbrace{(-a)+(-a)+\ldots+(-a)}_{n \text{ times }}=n(-a)=-na$

$n\in\mathbb{Z}$ because $n$ is the number of $a$'s in the element $na$. In the case $n<0$, $n$ is the number of $-a$'s in the element $na$.