From Pinter 22.G.6

Let $G$ be a group, $a, b \in G$.

Let $S = \{n \in \mathbb{Z} : ab^n = b^na\}$

Prove that $S$ is an ideal of $\mathbb{Z}$.

For an abelian group, $S = \mathbb{Z}$. For a non-abelian group, $S$ will contain all the multiples of the group order. But I don't know how to determine if there would (or wouldn't) be other values.

  • $\begingroup$ Think about a dihedral group, say the symmetry group of a regular polygon. Let $a$ be a generator of the rotations, and $b$ a reflection. What is $S$ here? $\endgroup$ Sep 15 '17 at 21:07
  • $\begingroup$ If $ab^n = b^n a, a b^m =b^m a$ then $a b^{n+m} = ?$. More generally $C_a = \{ g \in G, ag = ga\}$ is a subgroup of $G$ . $\endgroup$
    – reuns
    Sep 15 '17 at 21:31
  • 1
    $\begingroup$ The ideals of $\mathbb{Z}$ coincide with the (additive) subgroups; this greatly simplifies the proof. $\endgroup$
    – egreg
    Sep 15 '17 at 21:44
  • 1
    $\begingroup$ For fixed $b$, $n \cdot x := b^{-n} x b^n$ gives a group action of $\mathbb{Z}$ on $G$, and then $S$ is the stabilizer of $a$ under this group action. $\endgroup$ Sep 15 '17 at 22:32

First of all, we see that

$ab^0 = a = b^0a = a, \tag 1$

so $0 \in S$, always, for any $a, b \in G$; $S \ne \emptyset$, ever, and if $S = \{0\}$, we are done. Next, we observe that if $n \in S$, $-n \in S$ also, since

$ab^n = b^na \Longleftrightarrow b^{-n}ab^n = a \Longleftrightarrow b^{-n}a = ab^{-n}; \tag2$

thus if $S \ne \{0\}$, it contains a positive integer $m$; as such, $S$ contains a least positive element $p$; now for any $t \in S$, we may by the division algorithm write

$t = pq + r, \tag 3$

with either $r = 0$ or $1 \le r < p$; since $t \in S$,

$b^t a = ab^t, \tag 4$


$b^{pq + r} a = ab^{pq +r}; \tag 5$

we have:

$b^{p(q - 1)+ r} b^p a = b^{pq + r} a = ab^{pq + r}, \tag 6$

but since

$b^p a = ab^p, \tag 7$

(6) yields

$b^{p(q - 1)+ r} a b^p = ab^{pq + r}, \tag 8$


$b^{p(q - 1) + r} a = a b^{(q - 1) + r}; \tag 9$

we can continue in this manner until we reach

$b^r a = a b^r; \tag{10}$

but this contradicts the minimality of $p$ unless $r = 0$; thus $t = pq$ is a multiple of $p$; it is clear from (7) that every multiple of $p$ is in $S$; so

$S = (p), \tag{11}$

the principal ideal in $\Bbb Z$ generated by $p$.


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