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Bijection of a group

  • Thread starter Gale
  • Start date
  • #1
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1. The problem statement, all variables and given/known data
Let (G,*) be a group, and denote the inverse of an element x
by x'. Show that f: G to G defi ned by f(x) = x' is a bijection,
by explicitly writing down an inverse. Given x, y in G, what is
f(x *y)?

Homework Equations





The Attempt at a Solution



Okay, I think I'm just over thinking this... because it seems obvious that f is a bijection, since any function that has an inverse is bijective, and f obviously has an inverse where f-1(x')=x. And since x' and x are both in G, and G is a group then f-1 is well defined. But I keep having trouble with how to write proofs like this. Could someone help me out? Thanks!
 

Answers and Replies

  • #2
jgens
Gold Member
1,580
49
Try writing something like: "Notice that [itex]f \circ f = \mathrm{id}_G[/itex] and this implies that [itex]f[/itex] is a bijection."
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,777
911
Yes, since the inverse of the inverse of a group element is the group element itself so this function is its own inverse.
 

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