0

Suppose G is a group that has exactly eight elements of order 3. How many subgroups of order 3 does G have?.I know that the order of an element as well as the order of subgroup divides the order of the group. But I am not able to apply the logic here.

quid
  • 42,135
user378334
  • 45
  • Do not completely alter your questions! Ask a new one instead. – quid Oct 20 '16 at 18:14
  • Stop editing this question to ask new ones. – Tobias Kildetoft Oct 20 '16 at 18:15
  • I want to do the same but the site is not accepting my questions. I am a new user on this site and one of my question was heavily down voted. What should I do so that I can ask new questions again please suggest. – user378334 Oct 20 '16 at 18:19
  • It's a temporary block. At the moment, you can only wait until it expires next week. To avoid further blocks, provide some context to your questions in future. Questions that consist of a bare problem statement, like this one tend to attract downvotes. If you had for example said something about what you know of computing the radius of convergence of a power series (do you know the ratio test or the root test? the Cauchy-Hadamard formula?) it would be less likely to attract downvotes. – Daniel Fischer Oct 20 '16 at 18:46

1 Answers1

3

The mapping $g\mapsto \langle g\rangle$ from elements of order $3$ to the subgroups they generate is a 2-to-1 mapping from the set of elements of order $3$ onto the set of subgroups of order $3$. So, if you have a 2-to-1 mapping of an 8-element set onto a second set, how big must the second set be?

Edit: I respond to a comment:

If $g$ has order $3$, then $\langle g\rangle = \{1, g, g^{-1}\}$. Under the mapping that assigns an element $g$ to the subgroup it generates, both $g$ and $g^{-1}$ get assigned the subgroup $\langle g\rangle$, and these are the only elements that get assigned this subgroup. Hence the mapping is 2-to-1 from the set of elements of order 3 to the set of subgroups of order 3.

Keith Kearnes
  • 13,798