Exercise · ZW-GROUP-ACTION-008 · ★★☆ Intermediate
In an action of \(G\) on \(X\), what kind of function is \(x\mapsto g\cdot x\) from \(X\) to \(X\), for each \(g\in G\)?
Choose one option.
Show answer
- D. A bijection
Explanation
It has inverse \(x\mapsto g^{-1}\cdot x\), since \(g^{-1}\cdot(g\cdot x)=x\) and \(g\cdot(g^{-1}\cdot x)=x\). Hence it is a bijection.
Review the concept: Group action
An action of a group \(G\) on a set \(X\) assigns to each \(g,h\in G\) and \(x\in X\) an element \(g\cdot x\in X\), so that the identity \(e\) moves nothing and the product of \(G\) acts by composition.
\[e\cdot x=x,\qquad (gh)\cdot x=g\cdot(h\cdot x)\]
More practice problems: group action
- ZW-GROUP-ACTION-001 The group S_3 acts on X=\1,2,3\ by g· x=g(x). If g=(1\ 2\ 3), what is g·1?
- ZW-GROUP-ACTION-002 In an action of G on X, what is e· x for x∈ X?
- ZW-GROUP-ACTION-003 Which of the following formulas defines an action of (mathbb Z,+) on mathbb Z?
- ZW-GROUP-ACTION-004 Let S_3 act on \1,2,3\, with g=(1\ 2\ 3) and h=(1\ 2). Compute (gh)·1, composing from right to left.
- ZW-GROUP-ACTION-005 In S_3 acting on \1,2,3\, we have g·1=2 with g=(1\ 2\ 3). What is g^-1·2?
- ZW-GROUP-ACTION-006 The group mathbb Z_6 acts on itself by g· x=g+x (mod 6). Compute 4·5.
- ZW-GROUP-ACTION-007 Let S_3 act on itself by conjugation, g· x=gxg^-1. Compute (1\ 2)·(1\ 2\ 3).
- ZW-GROUP-ACTION-009 mathbb Z_6 acts on mathbb Z_6 by n· x=x+2n (mod 6). How many distinct values does n·0 take as n ran…