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 $k$? $\endgroup$
    – Asinomás
    Commented May 16, 2018 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
    Commented May 16, 2018 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
    Commented Jun 18, 2018 at 0:49

1 Answer 1


According to the definition, yes, you are correct so far, but your enumeration of the set as a set of 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.


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