2
$\begingroup$

Question: Let a be an element of a group and suppose that a has infinite order. How many generators does $\left \langle a \right \rangle$ have?

Following from the hypothesis that a has infinite order, distinct power on a are distinct group elements.

A bit of a brick wall here.

Only hints are appreciated unless it involves substantial number theory. Thanks in advance.

  • 0
    Said differently, for the subgroup $H = <1>$ of $(\mathbb{Z},+)$ what is the subgroup of $H$ generated by the element $1\,?$ – 2017-02-26
  • 0
    One more hint: What is the _definition_ (in words) of $<1>$? – 2017-02-26
  • 0
    $1 \in \mathbb{Z} so \left \langle 1 \right \rangle$ is a cyclic subgroup of the group $\left ( \mathbb{Z},+ \right )$. In fact, $\left \langle 1 \right \rangle = \left ( \mathbb{Z},+ \right ) $so$ \left ( \mathbb{Z},+ \right )$ is a cyclic group. To determine the generators of cyclic groups we recall: G=$\left \langle a^{k} \right \rangle$ IFF$ gcd\left ( n,k \right )=1$ @bof – 2017-02-26
  • 0
    @quasi $\left \langle 1 \right \rangle$ is a cyclic subgroup generated by the element 1 in $\mathbb{Z}$ – 2017-02-26
  • 1
    Ok, so is the element $1$ a generator of $(\mathbb{Z},+)$? – 2017-02-26
  • 0
    @quasi of course it is. Observe: $\left \langle 1 \right \rangle = \left \{ 1^{n} : n \in \mathbb{Z}\right \}=\left \{\cdot \cdot \cdot ,-3,-2,-1 ,e, 1, 2, 3,\cdot \cdot \cdot \right \}=\left ( \mathbb{Z},+ \right )$ – 2017-02-26
  • 1
    Is the element $2$ a generator of $(\mathbb{Z},+)$? – 2017-02-26
  • 0
    It is not. Observe: $3 \in \mathbb{Z} but \left \langle 2 \right \rangle = \left \{\cdot \cdot \cdot , -6,-4,-2,e,+2,+4,+6,\cdot \cdot \cdot \right \}\nsupseteq 3$@quasi – 2017-02-26
  • 1
    Is there any other integer that will generate $(\mathbb{Z},+)$? – 2017-02-26
  • 0
    @quasi -1 would. In fact, there are two generators; 1 and -1. – 2017-02-26
  • 0
    Now you got it. – 2017-02-26
  • 0
    A notational comment: For $(\mathbb{Z},+)$, the identity element is $0$, so calling it $e$ is not necessary (and not usually done). – 2017-02-26