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The book:

The notation ka in additive notation does not represent a product of k and a but, rather, a sum $ka=a+a+a+\cdots+a$ with k terms

I confused about the above definition, because we say that $(-1)$ is generator for integers,but we need negative $k$ values to obtain positive integers. If $ka$ is not product, then how can we cancel out the negative sign for getting positive integers ?

I am beginner at abstract algebra and not native speaker. Maybe my question seems stupid, but I really did not understand what it is wanted to mean by saying "The notation ka in additive notation does not represent a product of k and a",because if it is not product, then we cannot obtain the positives

Arturo Magidin
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  • This seems indeed confusing, but may I ask, it is a generator in what sense? As a ring, or just an abelian group?Maybe this is just to pinpoint the fact that this notation can be used in any abelian group and before being a product, it I a sum. It is a product iff you have integers in it – julio_es_sui_glace Mar 06 '24 at 19:34
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    $ka$ is only a "short form" for the addition $a+a+\cdots +a$, because we only consider the group $(\Bbb Z,+)$, which does not have a multiplication, only addition. And we can say $k(-1)=(-1)+(-1)+\cdots +(-1)$, so I don't see any problem. – Dietrich Burde Mar 06 '24 at 19:47
  • @DietrichBurde how do you obtain $+3$ for example using $(-1)$ –  Mar 06 '24 at 19:49
  • @DietrichBurde if it is just a short form, then how can it also be generator of Z –  Mar 06 '24 at 19:49
  • $g$ is a generator means, that every element $x$ is a "power" of $g$, so $x=g^n$ with $n\in \Bbb Z$. For addition instead of multiplication it says that $x=kg$ with $k\in \Bbb Z$. So we only need to adapt the notation $ka$ for negative $k$ accordingly. So for $k=-3$ and $a=-1$ it is $ka=1+1+1=3$. The $g^{-1}$ becomes $-g$ for addition. – Dietrich Burde Mar 06 '24 at 19:51
  • @DietrichBurde yes, but when we adapt it for negative k values, then there would be product of k and g , but the book says we cannot say "product" in definition, so I confused in putting negative k and obtain positive number –  Mar 06 '24 at 19:55
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    No, it is never a product. For $k=3$ and $a=1$ is is $ka=1+1+1=3$ and for $k=-3$ and $a=-1$ it is $ka=-(-1)+-(-1)+-(-1)=3$. The generators of $(\Bbb Z,+)$ are only $a=1$ and $a=-1$. So we just want to abbreviate the addition of $k$ terms $1$, or $k$ terms $-1$ for $k\in \Bbb Z$ by writing a "fake product" $ka$. Everything gets much easier if you pass to the multiplicative version $(C_{\infty},\cdot)={e,g^{\pm 1},g^{\pm 2},\ldots }$. – Dietrich Burde Mar 06 '24 at 19:57
  • maybe just complete your notation by : if $k\in \mathbb Z$ and $k<0$, then the notation $ka$ means $(-k)(-a)$ – Stéphane Jaouen Mar 06 '24 at 20:07
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    $0 = 0a = (-k+k)a = (-k)a+ka ,\Rightarrow, \color{#c00}{(-k)a = -(ka)},,$ i.e. $,(-k)a,$ is the additive inverse of $,ka.\ $ Written in multiplicative notation we have: $\ \color{#c00}{a^{-k} = (a^k)^{-1}}\ \ $ – Bill Dubuque Mar 06 '24 at 20:08
  • @BillDubuque : What allows you to use distributivity if you're just in an additive group? – Stéphane Jaouen Mar 06 '24 at 20:13
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    @Stéphane $,(j+k)a = ja+ka,$ and $,a^{j+k}=a^j a^k,$ are true because addition and multiplication are associative in the respective groups (so we can drop parentheses) $\ \ $ – Bill Dubuque Mar 06 '24 at 20:27
  • @Bill : Provided that you have previously given meaning to $g^{k}:=(g^{-1})^{-k}$ for $k<0$. And THEN you prove that you can drop parentheses. Isn't it ? Otherwise, there is a parallogism. – Stéphane Jaouen Mar 06 '24 at 20:35
  • Which book are you referring to? – Shaun Mar 07 '24 at 02:37

1 Answers1

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To gather up the several good comments, etc.:

First, there are at least two definitions/characterizations of "subgroup $H$ of group $G$ generated by a subset $S$ of $G$". The slightly fancy one, but whose utility we can eventually see, is to take $H$ to be the intersection of all subgroups of $G$ containing $S$. (And prove that arbitrary intersections of subgroups are, again, subgroups.) This characterization skirts the issue in the question.

Second, define the subgroup-generated-by $S$ to be the collection of all "words" in $S$... and in their inverses! Otherwise, it won't be a subgroup.

In the simplest case of the latter, where $S=\{a\}$ as in the original question, it's not just $a+\ldots+a$ expressions, but, also, $(-a)+\ldots+(-a)$ expressions (and $0$, the "empty sum", by (reasonable) convention).

The issues about $k\cdot a$ and $(-k)\cdot a$ and so on are about notation, not so much about mathematical facts. Yes, there is indeed some checking to be done to be sure that everything is well-defined, and that presumed properties really do hold (depending, of course, on the choice of logical development of these basics...) Yes, this checking is not completely trivial, although it's easy. :)

paul garrett
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