From Abstract Algebra in GAP by Alexander Hulpke, section 3.23 "Groups generated by elements":

We have seen so far two ways of specifying subgroups: By listing explicitly all elements, or by specifying a defining property of the elements. In some cases neither variant is satisfactory to specify certain subgroups, and it is preferrable to specify a subgroup by generating elements.

Definition: Let $G$ be a group and $a_1, a_2, ..., a_n ∈ G$. The set

$<a_1,a_2,\ldots,a_n> = \\ \bigg \{ b_1^{\epsilon_1} \cdot b_2^{\epsilon_2} \ldots \ldots b_k^{\epsilon_k} | k \in N_0 = \{0,1,2,3,\ldots\}, b_i \in \{a_1, \ldots, a_n \}, \epsilon_i \in \{ 1, -1 \} \bigg \} $

(with the convention that for $k = 0$ the product over the empty list is $e$) is called the subgroup of $G$ generated by $a_1,...,a_n$.

My question is around the interpretation of the set notation above.

Let's say we have <a,b>. If I'd like to enumerate the set according to the definition above, is this the correct interpretation?

k = 0    ()

k = 1


k = 2    

    a    a
    a    a^-1
    a^-1 a
    a^-1 a^-1

    a    b
    a    b^-1
    a^-1 b
    a^-1 b^-1

    b    a
    b    a^-1
    b^-1 a
    b^-1 a^-1

    b    b
    b    b^-1
    b^-1 b
    b^-1 b^-1

k = 3

  • 1
    $\begingroup$ yes, that seems to be the correct interpretation, although its hard to tell only with $k=0,1,2$. How do you generate it for larger $k$? $\endgroup$ – Jorge Fernández Hidalgo May 16 '18 at 3:15
  • 1
    $\begingroup$ @dharmatech: You have the right idea. For general $k$, we want all possible products of $k$ terms, where each term is either $a_i$ or $a_i^{-1}$, for some $i$. Hence in your example with two generators $a,b$, each of the $k$ factors must be one of $a,a^{-1},b,b^{-1}$. $\endgroup$ – quasi May 16 '18 at 3:19
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$ – Shaun Jun 18 '18 at 0:49

According to the definition, yes, you are correct so far, but your enumeration of the set as a set of a group elements is flawed in that it fails to take into account that cancellation happens each time an element and its inverse occur next to each other.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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