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Let $(G,\cdot)$ be a group and $a\in G$.

Let's consider the set $\{a, ... , a^{|G|} \}$. Imagine the elements of that set repeat themselves (say, $4$ times) then this will imply that there are $4$ $i$'s in this set such that $a^i = 1$ (the reason is because $a^r = a^s$ with $s>r \implies a^{r-s}=1$). Let $n= \min \{i | a^i=1\}$.

My question is how can we be sure that there will not be any repetition of elements in the set $\{a, ..., a^n \}$?

The context of this question is to understand why does in this answer it is said that $n$ is the order of the group $\{a, ..., a^n \}$, where $n$ is defined as it was above.

Many thanks!

EDIT: I think I've got it: the corresponding $1$ will always be either on the left or in-between the pair of repeated elements?

Shaun
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niobium
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  • If there were $a^r$ and $a^s$ in ${a,\dots,a^n}$ that are equal, then taking $s>r$ we have $$a^r=a^s$$ $$\implies 1=a^{s-r}$$ and $s-r$ is less than $n$. Which contradicts $n$ being the smallest $i$ such that $a^i=1$ – Zoe Allen Nov 30 '23 at 10:59

1 Answers1

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Assume there were $a^i=a^j$, $i,j\leq n$. Without loss of generality, we can assume $i\leq j$.
This implies, like you already wrote, $a^{j-i}=1$, so, by choice of n, either $j-i=0$ or $j-i = n$.
$j-i=n$ can't be by choice of $i,j$, so $i=j$, which means there is no repition in the set you provided (Specifically, the set has cardinality $n$).

Nicky Hekster
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watertrainer
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