# Number of cyclic subgroups of order $10$ in $\mathbb{Z}_{30}\oplus\mathbb{Z}_{120}$.

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.

• $14$ is wrong , since the number of elements of order $10$ is multiple of $\phi(10)=4$ – Chinnapparaj R Apr 20 '19 at 3:46
• 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. – Rylee Lyman Apr 20 '19 at 3:49
• 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. – Rylee Lyman Apr 20 '19 at 4:14
• The operation is addition. – AColoredReptile Apr 20 '19 at 4:21

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 $$x$$

• $\phi(4)=2$ so should the final answer be $28$? – Ross Millikan Apr 20 '19 at 4:01
• $(1,10)$ does not have order $10$, since $10\cdot(1,10) = (10,100) \ne (0,0)$! – Rylee Lyman Apr 20 '19 at 4:02
• sorry! its $\phi(10)=4$ – Chinnapparaj R Apr 20 '19 at 4:02
• $|b|$ means order of b – Chinnapparaj R Apr 20 '19 at 4:06
• What about $|a|=|b|=10$? This adds an extra $16$ elements of order $10$. – Theo C. Apr 20 '19 at 4:40

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.