Order of elements in $\mathbb{Z}_{27}$

Question

When calculating orders of elements in the groups $\mathbb{Z}_n$ this was my thinking:

For $\mathbb{Z}_{15}$, I would do $|3|=5$ because $3$ needs to be multiplied by $5$ to give a multiple of $15$ and so on with the rest of the elements. This worked up to $\mathbb{Z}_{15}$, but when calculating the order of the elements in $\mathbb{Z}_{27}$ it doesn't work. Could someone explain why this is? And maybe an easier way of calculating the order for the elements in $\mathbb{Z}_n$.

The order of the element $(3, 3)$ in the group $\mathbb{Z}_{9} × \mathbb{Z}_{27}$ gives $9$ which implies that both orders are $3$, so my calculation of the order of $3$ in $\mathbb{Z}_{27}$ is wrong ( I thought it was $9$).

Answer 1

The order of $3$ is $\mathbb{Z}_{27}$ is $9$. If the order were $3$, $(3,3)$ would have order $3$ in $\mathbb{Z}_9\times \mathbb{Z}_{27}$, not order $9$. The order of an element of a direct product is the least common multiple of the orders of its components, not the product.

Answer 2

Since $\mathbb{Z}_{n}$ is a cyclic group, the number of elements of order $d$ where $d | n$, is $\phi(d)$, you can find some reference here

($\phi$ denoted the Euler function)

Answer 3

Actually in order to find the order of an element $a$ in a cyclic groups like this $\mathbb {Z}_{27}$ you need to think more like if $|a|=k$, then $k$ is the least positive integer such that if you multiply $a$ with $k$ then you'll get $27$ or any multiple $27$.

Answer 4

One thing to add to the other answers, the reason the number of elements of order $d$, where $d|n$, is $\phi(d)$, is the basic fact about cyclic groups: $|a|=n\implies |a^k|=n/\operatorname{gcd}(n,k)$. The proof of this fact is a good exercise, though quite straight forward.