If $a$ is of order $n$, then $a^i = a^j$ if and only if $i \equiv j \pmod {n}$.

Question

Let $(G, *)$ be a group. Let $a$ be an element in the group.

If $a$ is of order $n$, then $a^i = a^j$ if and only if $i \equiv j \pmod {n}$.

I don't understand conjugate classes in groups; which is usually used to prove such problems, as here (though it is a different question there, and not found this question here), though understand that both an element and it's inverse have the same order in a group.

But, the intuitive meaning is seemingly easy, i.e. if apply group operation (let $\circ$) $i$ times to an element $a$; then the map is same to $a^j$, iff $\overline{i}=\overline{j}$.

But, why need more complicated concepts like: conjugate class concept, or conjugation?
Request logic. Also if possible provide a solution using conjugation, with reference to a small example say : let $(G, *)$ be given by:$\langle \mathbb{Z}/ \mathbb{7Z},+\rangle.$

Answer 1

Suppose $a^i=a^j$. Then $a^ia^{n-i}=a^ja^{n-i}\implies a^n=a^{n+j-i}\implies 1=a^{n+j-i}$

But we know that the order of $a$ is $n$, so we must have $n\mid n+j-i\implies n\mid j-i$. This proves one direction. The other direction is clear.

Answer 2

Unless you want to illustrate the use of a particular concept, such as conjugate classes, I have always tried to use the simplest techniques possible. This may, or may not, provide what you want but here goes. Suppose for the sake of argument that $i > j$ (the result is trivial if they are equal) so that $i-j$ is positive. Since $n$ is the least positive integer for which $a^t=e,$ use the division algorithm to write $i-j=qn+r$ with $0 \leq r \leq n-1.$ Since $a^{i-j}=a^r=e,$ we have a contradiction unless $r=0.$ Thus, $i-j$ is a multiple of $n$ and $i$ is congruent to $j$ mod $n.$ But, you probably already know this, so I apologize if this answer seems a bit pedantic.