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Let me first quote an example from Herstein text $(2$nd Ed.$)$:

Example 2.2.4 Let $n$ be any integer.We construct a group of order $n$ as follows: $G$ will consist of all symbols $a^i,i=0,1,...,n-1$ where we insist that $a^0=a^n=e,a^i.a^j=a^{i+j}$ if $i+j\leq n$ and $a^i.a^j=a^{i+j-n}$ if $i+j>n.$ The reader may verify that this is a group. It is called a cyclic group of order $n.$

Now my questions are:

How does $G$ look like if $n<0?$ Does for $n=-1,G=\{a^0,a^{-1},a^{-2}\}?$ But then how do $(a^{-2})^2=a^{-4}$ where $-4<-1$ can be calculated from the definition?

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Herstein meant positive integer, of course there is no group of order $-47$. But formally, for your $n=-1$, recall that the definition says $a^n=e$, so $a^{-1}=e$, everything is $e$. Everything works out fine, if we use $n=-24$ we get the cyclic group of order $24$. –  André Nicolas Jan 31 '13 at 16:46
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The definition is also problematic for $n=0$. It looks like a typo. –  sdcvvc Jan 31 '13 at 16:52

3 Answers 3

up vote 5 down vote accepted

$n$ should be taken as a positive integer there as we are talking about a group having order $n$. Note that in any group, we can talk about $a^{-n}$ which may be taken as $(a^{-1})^n=(a^n)^{-1}$.

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Maybe you are confusing the representation of elements with the group's operation they have; remember: a group must be closed under operation. Besides remember the definition for a order of a group: the smaller positive integer such that, for $g \in G$, $g^{n}=e$ the identity of $G$.

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Surely Herstein meant positive $\rm\,n.\:$ However, one can trivially alter the definition to also work with negative exponents. In fact we can let the exponents be any set $\rm\,S\,$ that is a complete system of reps for the congruence classes of the additive group $\rm\,\Bbb Z/n =$ integers mod $\rm\,n.\:$ For example, for $\rm\,n=5,\,$ we could let $\rm\,S\,$ be $\:0,-1,-2,-3,-4\ $ or $\ 2,1,0,-1,-2.\:$ Then the group operation is simply $\rm\:a^i a^j = a^{(i+j)\ mod\ n}\,$ where $\rm\:(i\!+\!j)\ mod\ n\:$ denotes the unique element of $\rm\,S\,$ that is $\rm\equiv i+j\,\ (mod\ n).$

This amounts to simply rewriting said additive group in a multiplicative form. If you know about group isomorphisms then you should find it an easy exercise to show that the multiplicative representation is isomorphic to the additive one. Indeed, the definition of exponent addition is precisely what is required to make that the map $\rm\:k\to a^k\:$ be a group homomorphism.

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