Showing that Aut(G) is a group

  • #1
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Homework Statement


Prove that, for any group ##G##, the set ##\operatorname{Aut} (G)## is a group under composition of functions.

Homework Equations




The Attempt at a Solution


1) associativity: It is a known fact of set theory that composition of functions is an associative binary operation.

2) identity: ##\operatorname{id}_G## is in ##\operatorname{Aut} (G)## because it is a group isomorphism from ##G## to ##G##. By the properties of the identity set-theoretic map, if ##f \in \operatorname{Aut} (G)## then ##\operatorname{id}_G \circ f = f \circ \operatorname{id}_G = f##.

3) inverse: Let ##f \in \operatorname{Aut} (G)##. Then the set-theoretic inverse ##f^{-1}## is still an isomorphism from ##G## to ##G## (I have proven this fact previously), so ##f^{-1} \in \operatorname{Aut} (G)##. And since ##f^{-1} \circ f = f \circ f^{-1} = \operatorname{id}_G##, every element in ##\operatorname{Aut} (G)## has an inverse.
 

Answers and Replies

  • #2
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11,685

Homework Statement


Prove that, for any group ##G##, the set ##\operatorname{Aut} (G)## is a group under composition of functions.

Homework Equations




The Attempt at a Solution


1) associativity: It is a known fact of set theory that composition of functions is an associative binary operation.

2) identity: ##\operatorname{id}_G## is in ##\operatorname{Aut} (G)## because it is a group isomorphism from ##G## to ##G##. By the properties of the identity set-theoretic map, if ##f \in \operatorname{Aut} (G)## then ##\operatorname{id}_G \circ f = f \circ \operatorname{id}_G = f##.

3) inverse: Let ##f \in \operatorname{Aut} (G)##. Then the set-theoretic inverse ##f^{-1}## is still an isomorphism from ##G## to ##G## (I have proven this fact previously), so ##f^{-1} \in \operatorname{Aut} (G)##. And since ##f^{-1} \circ f = f \circ f^{-1} = \operatorname{id}_G##, every element in ##\operatorname{Aut} (G)## has an inverse.
Correct, but you have forgotten what most students tend to forget. It's as easy as the rest, but for the sake of completion, it is very important: closure! Why is a composition of automorphisms an automorphism again?
 
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  • #3
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Correct, but you have forgotten what most students tend to forget. It's as easy as the rest, but for the sake of completion, it is very important: closure! Why is a composition of automorphisms an automorphism again?
4) closure: Let ##f,g \in \operatorname{Aut} (G)##. Then ##h = f \circ g \in \operatorname{Aut} (G)## becuase ##h## is still a bijection, and ##h## is a homomorphism because composition of homomorphisms results in a homomorphism: ##h (gg') = f(g(hh')) = f(g(h)g(h')) = f(h(g))f(h(g')) = h(g)h(g')##.
 
  • #4
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11,685
4) closure: Let ##f,g \in \operatorname{Aut} (G)##. Then ##h = f \circ g \in \operatorname{Aut} (G)## becuase ##h## is still a bijection, and ##h## is a homomorphism because composition of homomorphisms results in a homomorphism: ##h (gg') = f(g(hh')) = f(g(h)g(h')) = f(h(g))f(h(g')) = h(g)h(g')##.
I just type this to control your line. There are a bit too many ##g,h## meaning different things.
##h(ab)=f(g(ab))=f(g(a)g(b))= f(g(a))f(g(b))=h(a)h(b)##
 
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