Order of an element in a group G Suppose that $a$ is an element of order $n$ in a group $G$. Prove:
i) $a^i = a^j$ if and only if $i \equiv j \pmod n$;
ii) if $d = (m,n)$, then the order of $a^m$ is $n/d$;
I was trying to self teach myself this and came to this question. How would you solve this? Can someone please show how to?

If $i \equiv j \pmod n$, then $i = j + kn$ for some $k \in \Bbb Z$. It follows that:
$$
a^i = a^{j + kn} = a^ja^{kn} = a^j (a^n)^k = a^j
$$
If you haven't proved the power properties $a^{p+q} = a^pa^q$ and $a^{pq} = (a^p)^q$ for $p, q \in \Bbb Z^+$, this is a good exercise to do now. Try using induction.
Now, if $a^i = a^j$, then:
$$
a^i a^{-j} = a^{i - j} = 1
$$
And this is only possible if $i - j$ is a multiple of $n$. (i) follows.
The property $(a^n)^{-1} = a^{-n}$ was used here. Again, try proving it via induction.

For (ii), assume the order of $a^m$ is a number smaller than $n/d$, and try to use (i) to reach a contradiction.
