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How to find all the elements of order $8$ in the group ($\Bbb Z_{24}$, addition modulo $24$).

Order in the case of addition $\!\bmod{24}\,$ would mean, say an element $a$ has order $8$. Then $8a \bmod {24} = 0.$

From here, I can see that $3$ is one of the answers. Am I going about this wrong? I'm stuck here.

How do I find the elements without using any concepts related to subgroups?

Bill Dubuque
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ali
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    $8a\equiv0\bmod24\iff24|8a\iff 3|a$ – J. W. Tanner Feb 16 '20 at 17:15
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    Yes, 3 has order 8 in the group $\mathbb Z_{24}$ under modular arithmetic. What about $9$? $8\times 9= 72 \equiv 0\mod {24}$. But check that $6$ doesn't qualify, nor $12$. – amWhy Feb 16 '20 at 17:15
  • @J.W.T But $,8a\equiv 0,$ only implies $,o(a)\mid 8.,$ To get $,o(a) = 8,$ we have to also exclude the cases where $,o(a),$ is a proper factor of $,8,,$ e.g. see my answer. – Bill Dubuque Feb 16 '20 at 18:34
  • Of course you are correct, @BillDubuque; I merely made a comment to help OP get unstuck – J. W. Tanner Feb 16 '20 at 18:40

3 Answers3

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The order of $m$ in $\mathbb{Z}_n$ is $n/\mathrm{gcd}(m,n)$. Hence, in order to find the elements of order $8$ in $\mathbb{Z}_{24}$, it suffices to find the nonnegative integers less than $24$ whose gcd with $24$ is $24/8=3$. Those all precisely the positive odd multiples of $3$ less than $24$, which are $3$, $9$, $15$, and $21$.

Geoffrey Trang
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$o(x) = 8\iff 0\equiv 8x,\ 0\not\equiv 4x,\ $ by the Order Test
$\qquad\quad\ \ \begin{align} &\iff 24\mid 8x^{\phantom{|^|}}\!\!,\ 24\nmid 4x\\[.2em] &\iff \ \ 3\mid x,\ \ \ \ \ 6\nmid x,\ \ \ {\rm by \ cancelling}\, \ 8,\,4,\ \, \rm resp.\end{align}$

Bill Dubuque
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Yes, $3$ has order $8$ in $\mathbb Z_{24}$, and generates the subgroup of order $8$ given by $\langle 3\rangle = \{0, 3, 6, 9, 12, 15, 18, 21\}$.

Hint: In addition to $3$ having order $8$, confirm that so do the elements $9$ ($8\times 9=72 \equiv 0 \pmod 24)$; $15 (8\times 15= 120 \equiv 0\pmod{24})$, and I'll let you confirm that $21$ has order 8 in $\mathbb Z_{24}$ under modulo addition.

Each of them are generators of $\langle 3 \rangle = \{0, 3, 6, 9, 12, 15, 18, 21\}< \mathbb Z_{24}$ under addition, and hence are each of order $8$.

amWhy
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    The group generated by $3$ is of order 8. I never claim that all elements in $\langle 3 \rangle$ are of order 8. Only $3, 9, 15, and 21 are of order 8. Please read my answer far more carefully, @Squishy – amWhy Feb 16 '20 at 18:43
  • So if I had to find all elements that have the order 8, I would have to look at the elements of the subgroup generated by one of those elements (in this case: 3) ? – ali Feb 18 '20 at 06:28