Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to the introductory abstract algebra, it says that $g^n =e$ where $g$ is an element of some finite group. Then, it talks about the subgroup $\langle g^d \rangle$. What is it exactly, and what would be the elements of this group? (where $d$ is some integer.)

share|cite|improve this question
@William modified $n$ to $d$. – Lucy Zeo Aug 18 '12 at 1:20
up vote 1 down vote accepted

The notation $\langle g^d \rangle$ denotes the cyclic group generated by the element $g^d$. More explicitly, we begin with the element $g^d$. Then we consider $g^d \cdot g^d = g^{2d}$. We continue generating more elements of the subgroup and have the set $\{g^d, g^{2d}, g^{3d}, \cdots\}$ with the group operation coming from the finite group.

Of course it requires proof that this set generates a group. I will leave this as a fun exercise for you, but please ask if you need more help.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.