$o(z) = 28$, what is $o(z^{16})$?

Question

I have this math question that I am kind of stuck on.

Suppose that $o(z) = 28$, what is the order of the element $z^{16}$?

From this information I know that $o(z) = 28$ means that $z^{28} = 1$... How can I use that to find $o(z^{16})$? Thanks.

Answer 1

The fact that $o(z)=28$ gives you that $z^n = e$ if and only if $n = 28k$ for some $k\in {0,1,2,...}$, agreed? So you need to find the smallest $m$ such that $16m$ is divisible by $28$, i.e. least common multiple of $28$ and $16$ divided by $16$.

Answer 2

If $o(x)=n$, then $o(x^m)=o(x^d)$ where $d=(n,m)$.

Answer 3

The order of $u = z^4$ is $7$, and you're asked to find the order of $u^4$. The group generated by $u$ has prime order, so it is cyclic. As a result, all of its powers $u^k$ for $1 \leq k \leq 6$ will generate the whole group. Hence $u^4$ has order $7$ too.