Is it true that $a$ and $b$ should be disjoint permutations?

Question

Let $a,b \in S_n$ and $ab=ba$ and $b$ moves some points that not moved by $a$. Is it true that $a$ and $b$ should be disjoint permutations?

EDIT: We can consider $b=a^kc$ where $a$ and $c$ are disjoint, to construct a counterexample class. Do you know other constructions?

make a search about "centrilizer of an element in $S_n$", you will find what you want — mesel, May 25 '14 at 11:39

score 2 · Answer 1 · answered May 25 '14 at 07:04

2

No. Take $a=(1,2)$ and $b=(1,2)(3,4)$.

answered May 25 '14 at 07:04

Emanuele Paolini

score 2 · Answer 2 · answered May 25 '14 at 07:04

2

Not sure what you mean by ``disjoint permutation", but maybe $a=(1,2)$ and $b=(1,2)(3,4)$ (in $S_4$ for example) is a counterexample?

answered May 25 '14 at 07:04

DKal

score 2 · Answer 3 · answered May 25 '14 at 07:04

2

Consider a cycle $\,a\,$ and $\,b\,$ a product of $\,a\,$ with a cycle disjoint from it.

answered May 25 '14 at 07:04

anon

Excellent! We can consider $b=a^kc$ where $a$ and $c$ are disjoint, to construct more general class. Do you know other constructions? – May 25 '14 at 07:19
@user153200 Yes: $a$ and $b$ commute iff $b$ centralizes $a$, and we can describe the centralizer of a permutation explicitly in terms of its disjoint cycle representation. This gives not just any class of examples - it exhausts all of them. One small subclass of this is the permutations which can be written as products of powers of $a$'s cycles and any permutation disjoint from $a$. – anon May 25 '14 at 08:01

3 Answers3