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Determine the number of cyclic subgroups of order $10$ in $\mathbb{Z}_{30}\oplus\mathbb{Z}_{120}$.

So I want to find all the possible generators for these subgroups which means I want to find all the elements of order $10$ in the group and then take out all the elements which are duplicating subgroups.

I think I have $14$ elements of order $10$. I tried to count the elements such that $\text{lcm}(\vert x\vert,\vert y\vert)=10$ where $x\in \mathbb{Z}_{30},y\in \mathbb{Z}_{120}$. I have $\{(1,10),(2,20),...,(12,120),(5,2),(2,5)\}$. The first $12$ all generate the same subgroup as $\langle i\cdot(1,10)\rangle=\langle(i\cdot 1,i\cdot10)\rangle$.

So I have 3 distinct cyclic subgroups.

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  • $\begingroup$ $14$ is wrong , since the number of elements of order $10$ is multiple of $\phi(10)=4$ $\endgroup$ – Chinnapparaj R Apr 20 '19 at 3:46
  • $\begingroup$ In fact, implicit what you have shown is that $10\cdot(1,10) = (10,100) \neq (0,0)$, so $(1,10)$ does not have order 10. $\endgroup$ – Rylee Lyman Apr 20 '19 at 3:49
  • $\begingroup$ I'm not sure if this is helpful, but I want to ask you to be careful about whether your operation is multiplication or addition. If it is additon, then $(12,120) \equiv (12,0)$, say, is indeed an element of $\langle(1,10)\rangle$, and $1$ is not the identity. $\endgroup$ – Rylee Lyman Apr 20 '19 at 4:14
  • $\begingroup$ The operation is addition. $\endgroup$ – AColoredReptile Apr 20 '19 at 4:21
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Here $$10=|(a,b)|=\text{lcm}\{|a|,|b|\}$$ where $$|a| \in \{ 1,2,3,5,6,10,15,30\}$$ and $$|b| \in \{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\}$$

  • Case(1): $|a|=1,|b|=10$ (we have $4$ elements)
  • Case(2): $|a|=2,|b|=5$ (we have $4$ elements)
  • Case(3): $|a|=2,|b|=10$ (we have $4$ elements)
  • Case(4): $|a|=5,|b|=10$ (we have $16$ elements)
  • Case(5): $|a|=10,|b|=1$ (we have $4$ elements)
  • Case(6): $|a|=10,|b|=5$ (we have $16$ elements)
  • Case(7): $|a|=5,|b|=2$ (we have $4$ elements)
  • Case(8): $|a|=10,|b|=2$ (we have $4$ elements)
  • Case(9): $|a|=10,|b|=10$ (we have $16$ elements)

Thus totally we have $$2(4+4+4+16)+16=56+16=72$$ elements of order $10$ and hence we have $\frac{72}{\phi(10)=4}=18$ cyclic subgroups of order $10$


Notation: $|x|$ denotes the order of $x$

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  • $\begingroup$ $\phi(4)=2$ so should the final answer be $28$? $\endgroup$ – Ross Millikan Apr 20 '19 at 4:01
  • $\begingroup$ $(1,10)$ does not have order $10$, since $10\cdot(1,10) = (10,100) \ne (0,0)$! $\endgroup$ – Rylee Lyman Apr 20 '19 at 4:02
  • $\begingroup$ sorry! its $\phi(10)=4$ $\endgroup$ – Chinnapparaj R Apr 20 '19 at 4:02
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    $\begingroup$ $|b|$ means order of b $\endgroup$ – Chinnapparaj R Apr 20 '19 at 4:06
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    $\begingroup$ What about $|a|=|b|=10$? This adds an extra $16$ elements of order $10$. $\endgroup$ – Theo C. Apr 20 '19 at 4:40
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By the fundamental theorem of finite abelian groups, one has $$\mathbb{Z}_{30}\oplus\mathbb{Z}_{120} \simeq [\mathbb{Z}_5\oplus\mathbb{Z}_3\oplus\mathbb{Z}_2] \oplus [\mathbb{Z}_{5}\oplus \mathbb{Z}_3\oplus \mathbb{Z}_{8}] $$ Now, since each element in $\mathbb{Z}_3$ has order $3$ aside from the identity and $(3,10)=1$, I claim that this problem boils down to counting how many elements of order ten are in $G=\mathbb{Z}_2\oplus\mathbb{Z}_8\oplus\mathbb{Z}_5\oplus\mathbb{Z}_5$ (do you see why this holds true?).

Consider cases, letting $g=(a,b,c,d)\in G$ be arbitrary.

Case 1: $g=(1,b,c,d)$. Then $b$ can be either $0$ or $4$. If $c=0$ then $d\in \{1,2,3,4\}$, otherwise if $c\in \{1,2,3,4\}$ then $d\in\{0,1,2,3,4\}$ for a total of $2*(4+20)=48$ elements of order $10$.

Case 2: $g=(0,b,c,d)$. Then $b$ must be $4$. By a similar argument there are $24$ elements of order $10$ here.

We conclude by noting that $\frac{48+24}{\phi(10)}= \frac{72}{4} = 18$ cyclic subgroups of order 10.

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