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

Question

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

I have calculated that there are $24$ elements of order $10$ I know that in a cyclic subgroup of order $10$, There are $4$ element of order $10$

Thank you for helping.

Answer 1

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 ]) ]

Answer 2

You're almost there! You've got all the pieces you need; now it's just putting those pieces together.

Since you've correctly determined that there are $24$ elements of order $10$, and since you've correctly observed that every cyclic group of order $10$ contains $4$ elements of order $10$, then we need only divide: $$\text{There are $\;\dfrac{24}{4}$} = \text{$6\;$ cyclic subgroups of order $10\,$ in $\,\mathbb{Z}_{100}\oplus\mathbb{Z}_{25}$}$$