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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 $9$).

Matt Samuel
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MathGeek1998
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  • Which examples did not work in $\mathbb{Z}_{27}$? – lhf Apr 30 '20 at 17:53
  • the element $3$ in $\mathbb{Z}_{27}$, by doing it the way i stated above I get $|3|=9$, but in an exercise it says that it is supposed to be $3$ not $9$. – MathGeek1998 Apr 30 '20 at 18:02
  • Sarah, $\Bbb Z_n$ is actually a ring, right? So there are two different operations. The additive order of $3$ is, indeed, $9$. But since $(3,27)\gt1$, it is not a unit. I don't see how the order could be $3$. –  Apr 30 '20 at 18:26

4 Answers4

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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)

jacopoburelli
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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 $27$.

Itachi
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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.

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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.

Matt Samuel
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  • Ahh I see so, the individual orders are $9$ and $3$, but the order isn't the product of them, but the least common multiple which is $9$. – MathGeek1998 Apr 30 '20 at 20:01
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    @Sarah Yes, that's correct. – Matt Samuel Apr 30 '20 at 20:01
  • Thank you! I was confused with how to calculate the order of (a,b) in $\mathbb{Z}_n x \mathbb{Z}_m$ then. – MathGeek1998 Apr 30 '20 at 20:03
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    @Sarah You're welcome. The order in an additive abelian group is the smallest positive integer you can multiply the element by to get $0$. For an element of the direct product, you're multiplying both components by the same number, so the number has to be divisible by the orders of the two elements. Since you want the smallest positive integer, that is by definition the least common multiple. – Matt Samuel Apr 30 '20 at 20:05