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

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.

• Well, there should be 6 groups by what you say, right? – Prahlad Vaidyanathan Nov 3 '13 at 5:56
• I am not able to get how it will be $6$, and why? yes the answer is $6$ – Marso Nov 3 '13 at 6:02
• Every group will account for 4 elements of order 10, so the number of groups is $24/4 = 6$ – Prahlad Vaidyanathan Nov 3 '13 at 6:04
• Happy Diwali Sir! Thank you, It was quite easy, I was thinking two different cyclic group can have same element of order $10$ blah blah – Marso Nov 3 '13 at 6:07
• Happy Diwali :) – Prahlad Vaidyanathan Nov 3 '13 at 6:09

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 ]) ]

• Nice to use the latest GAP version at the time of answer! GAP 4.7.2 just become available! – Alexander Konovalov Dec 6 '13 at 16:27
• Looks like we should make a generic answer how to deal with questions like this in GAP, since this is really becoming a F.A.Q. ... – Alexander Konovalov Dec 6 '13 at 16:29
• @AlexanderKonovalov: Thanks for letting me know that availability. :-) – mrs Dec 6 '13 at 18:16

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: $$\text{There are \;\dfrac{24}{4}} = \text{6\; cyclic subgroups of order 10\, in \,\mathbb{Z}_{100}\oplus\mathbb{Z}_{25}}$$