Find all elements of order 6 in $(\mathbb{Z}_{48}, +)$

Question

The task is to find all element of order 6 in the group $(\mathbb{Z}_{48}, +)$.

I understand this task in a following way: we need to find all such $x \in \mathbb{Z}_{48}$, that $x \cdot 6 = 0\ mod\ 48$ As $id = 0$ for the mentioned group. And there is none $m$ less that 6, such as $x \cdot m = 0\ mod\ 48$. But as far as I understand $x^{0}=x \cdot 0 = id = 0$, so is it possible for any $x$ here to obtain order 6? I am sure that I've made mistake in my assumptions, as it would be odd for problem to have no answer. Do you have any ideas?

Answer 1

I'll stick to using additive notation in this answer (since we're considering $(\mathbb Z_{48}, +)$), but everything here works equally well if you use the multiplcative notation of $x^m$ instead of the additive $m\cdot x$.

The order of an element $x$ in a group is defined as the least positive integer $m$ such that $m\cdot x = 0$ where $m\cdot x$ is the addition of $x$ to itself $m$ times. While it's true that $0\cdot x = 0$, this does not mean that the order of $x$ is $0$ because $0$ is not positive. I will stick to, when we use this notation, putting $m$ (the integer) on the left and $x$ (the element of the group) on the right.

You can figure out that some $x$ in $(\mathbb Z_{48},+)$ satisfy $m\cdot x = 0$ exactly when $m\cdot x$ is, as an integer, a multiple of $48$. You would get that, for $m$ up to $6$, the solutions to $m\cdot x = 0$ are as follow: \begin{align*}&1\cdot 0 = 0\\ &2\cdot 0 = 0,\,2\cdot 24 = 0\\ &3\cdot 0 = 0,\,3\cdot 16 = 0,\,3\cdot 32= 0\\ &4\cdot 0 = 0,\,4\cdot 12 = 0,\,4\cdot 24= 0,\,4\cdot 36= 0\\ &5\cdot 0 = 0\\ &6\cdot 0 = 0,\,6\cdot 8 = 0,\,6\cdot 16 = 0,\,6\cdot 24 = 0,\,6\cdot 32, = 0,\,6\cdot 40 = 0\, \end{align*} You can then just take the numbers which are in the final list but no earlier list - which is just $8$ and $40$. This is essentially the most direct way to apply the given definition (though it is certainly not the most efficient way).

Answer 2

Since $\Bbb{Z}_{48}$ is cyclic, it has unique cyclic group of order $6$.

Clearly, $8$ is an element of order $6$ as $6(8)=0$.
As mentioned, every element of order $6$ must be in $\langle 8\rangle$.
Note that $|n8|=8$ iff $(n,6)=1$ iff $n=1$ or $5$.
So the other element of order $6$ will be $5(8)=40$.

Answer 3

You must decide: are going to use the $+$ notation for the group operation or the $.$ notation. Using both of them is a really bad idea.

I will use the $+$ notation. Asserting that $m$ has order $6$ means two things:

1. $m+m+m+m+m+m\equiv0\pmod{48}$;
2. if $k<6$, then $\overbrace{m+\cdots+m}^{k\text{ times}}\neq0$.

For instance, $16$ has order $6$, since $16+16+16+16+16+16+=96\equiv 0\pmod{48}$. But $5$ doesn't, since $5+5+5+5+5+5=30\not\equiv0\pmod{48}$.