1
$\begingroup$

If $G$ is a group and $a_i \in G$ for $i \in I$ where $I$ is a set of indices, then the subgroup $H$ of $G$ generated by $\{a_i\ |\ i\in I\}$ has as elements precisely those elements of $G$ that are finite products of integral powers of the $a_i$ where powers of a fixed $a_i$ may occur several times in the product.

I didn't understand what this theorem means. I didn't understand the part in bold. Does finite products of integral powers of the $a_i$ mean $a_i^k$ where $k$ is finite? Didn't understand the last part anyway, can someone somehow show a precise way of this theorem to me? Thanks!

1 Answers 1

2

This means that each element $g$ of the group can be written as $$g=a_{i_1}^{n_1}a_{i_2}^{n_2}\cdots a_{i_t}^{n_t},$$ where each $n_s$ are elements from $\Bbb Z$.

  • 0
    Might be as well to add that the $i_r$ don't all need to be different. – 2017-02-26
  • 0
    @MartinRattigan, yes, but $a_{i_r}$ and $a_{i_{r+2}}$ could be the same – 2017-02-27
  • 0
    So could $a_{i_r}$ and $a_{i_{r+1}}$, and $a_{i_r}$ and $a_{i_r}$ definitely will be. I think perhaps didn't notice the word "don't" in my previous comment. – 2017-02-28