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

group-theory

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.

Best Answer

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 $x$

Related Solutions

[Math] How Many Cyclic Subgroups of order $10$ are there in $\mathbb{Z}_{100}\oplus\mathbb{Z}_{25}$

By using GAP 4.6.5 these desired subgroup are enclosed computationally. Below is needed codes for doing this job:

gap> f:=FreeGroup("a","b");;
     a:=f.1;; b:=f.2;;
     s:=f/[a^100,b^25,a*b*a^(-1)*b^(-1)];;
     e:=AllSubgroups(s);;
     c:=Filtered(e,t->IsCyclic(t)=true);;
     Filtered(c,t->Order(t)=10);

 Group([ b^5, a^-50 ]),
 Group([ a^-9*b^10*a^-1 ]),
 Group([ a^-9*b^-10*a^-1 ]),
 Group([ a^8*b^-5*a^2 ]),
 Group([ a^8*b^5*a^2 ]), 
 Group([ a^-10 ]) ]

[Math] The number of cyclic subgroups of order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$

Provided you correctly counted the elements of order$~15$, your answer is correct. You can indeed count cyclic subgroups by counting their generators (elements or order$~n$) and dividing by the number $\phi(n)$ of generators per cyclic subgroup, since every element of order$~n$ lies in exactly one cyclic subgroup of order$~n$ (the one that it generates).

Here is how I would count the elements of order$~15$. By the Chinese remainder theorem one has $\newcommand\Z[1]{\Bbb Z_{#1}}\Z{30}\cong\Z2\oplus\Z3\oplus\Z5$ and $\Z{20}\cong\Z4\oplus\Z5$, so all in all we are dealing with the group $\Z2\oplus\Z4\oplus\Z3\oplus\Z5^2$. To have order $15$, an element must have a trivial (zero) component in $\Z2$ and $\Z4$, in $\Z3$ it must have as component one of the $2$ generators, and it's component in $\Z5^2$ must be any one of the $24$ nonzero elements. Indeed you get $2\times24=48$ elements of order$~15$.

Best Answer

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[Math] How Many Cyclic Subgroups of order $10$ are there in $\mathbb{Z}_{100}\oplus\mathbb{Z}_{25}$

[Math] The number of cyclic subgroups of order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$

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