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.

  • $\begingroup$ Well, there should be 6 groups by what you say, right? $\endgroup$ – Prahlad Vaidyanathan Nov 3 '13 at 5:56
  • $\begingroup$ I am not able to get how it will be $6$, and why? yes the answer is $6$ $\endgroup$ – Marso Nov 3 '13 at 6:02
  • $\begingroup$ Every group will account for 4 elements of order 10, so the number of groups is $24/4 = 6$ $\endgroup$ – Prahlad Vaidyanathan Nov 3 '13 at 6:04
  • $\begingroup$ 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 $\endgroup$ – Marso Nov 3 '13 at 6:07
  • $\begingroup$ Happy Diwali :) $\endgroup$ – 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;;

 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 ]) ]
  • $\begingroup$ Nice to use the latest GAP version at the time of answer! GAP 4.7.2 just become available! $\endgroup$ – Alexander Konovalov Dec 6 '13 at 16:27
  • $\begingroup$ 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. ... $\endgroup$ – Alexander Konovalov Dec 6 '13 at 16:29
  • $\begingroup$ @AlexanderKonovalov: Thanks for letting me know that availability. :-) $\endgroup$ – 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: $$\text{There are $\;\dfrac{24}{4}$} = \text{$6\;$ cyclic subgroups of order $10\,$ in $\,\mathbb{Z}_{100}\oplus\mathbb{Z}_{25}$}$$


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