An automorphism of a group G is an isomorphism from G to itself. Denote by...

Algotssleeddynf

Algotssleeddynf

Answered

2022-01-12

An automorphism of a group G is an isomorphism from G to itself. Denote by Aut G the set of all automorphisms of G.
a) Prove that Aut G is a group with respect to the operation of composition
b) Give an example of an abelian G such that Aut G is not abelian

Answer & Explanation

alkaholikd9

alkaholikd9

Expert

2022-01-13Added 37 answers

Step 1
Let GG be a group. By definition
Aut(G)={f:GGf is an isomorphism of G}.
Given two automorphisms f, gAut(G), we can consider the composition g×f. We claim that Aut(G), × is a group. We have to check all axioms.
First of all we need to show that g×f is again an automorphism, i.e. a homomorphism that is bijective. Now since g and f are bijective, g×f is bijective. Moreover,
(g×f)(ab)=g(f(ab))=g(f(a)f(b))=g(f(a))g(f(b))=(g×f)(a)(g×f)(b),
for all a, bG. Hence g×f is a group homomorphism.
Second, we must demonstrate that × is associative, i.e. (h×g)×f=h×(g×f). Just evaluate both morphisms at aG and see that both expressions coincide due to the associativity of GG.
Thirdly, we must confirm that a neutral component exists for ×. Clearly IdG:GG:aa is an automorphism. Since f×IdG=IdG×f for all fAut(G), IdG is the neutral element.
Last but not least, we must ensure that each fAut(G) has an inverse for ×. Consider the inverse function f1. Clearly f1×f=IdG=f×f1. So it remains to show that f1 is a group morphism. Now it's a very good exercise to prove this last statement.
EDIT: Let's prove the last statement. Suppose that f:GG is a group isomorphism. We need to show that f1 is a group morphism. Let a, bG. By definition there exist a unique x, yG such that f(x)=a and f(y)=b. Hence
f1(ab)=f1(f(x)f(y))=f1(f(xy))=xy.
Similarly
f1(a)f1(b)=f1(f(x))f1(f(y))=xy.
Therefore, f1(ab)=f1(a)f1(b)

Stella Calderon

Stella Calderon

Expert

2022-01-14Added 35 answers

Step 1
Let Aut(G) be the set of all automorphisms ϕ:GG. In order to show that this is a group under the operation of composition, we must verify:
1) Is the set is closed under composition? Yes! If you are given isomorphisms ϕ, ψ:GG, then it is not too tough to show that ψϕandϕψ are isomorphisms. I can expand on this in more detail if you like, but you have probably seen a proof before that a composition of bijective functions is bijective. If a and b are elements of the group,
ψϕ(ab)=π(ϕ(ab))=ψ(ϕ(a)ϕ(b)),
because ϕ is an isomorphism. Since ψ is also an isomorphism,
ψ(ϕ(a)ϕ(b))=ψϕ(a)ψϕ(b),
so the composition ψϕ preserves products. Thus, ψϕ is an isomorphism if ψ and ϕ are.
2) Is the set associative? Yes! All you need to do is show that, for any three isomorphisms ϕ, ψ and
ξ, ϕ(ψξ)=(ϕψ)ξ.
To do that, just show that for each x in
G,ϕ(ψξ)(x)=(ϕψ)ξ(x)=ϕ(ψ(ξ(x))).
It's just pushing around definitions.
3) Does the set contain an identity element? Yes! Let the identity automorphism e:GG be the map e(x)=x. Clearly, eϕ=ϕe=ϕ.
4) Does each element of the set have an inverse under ? Yes! Since each isomorphism ϕ:GG is bijective, there is a well-defined inverse map ϕ1:GG. You may have already seen a proof that the inverse of an isomorphism is an isomorphism. If not, it isn't too difficult to prove: I'll leave it to you, but I can expand on it if you need me to. Further, the composition
ϕ1ϕ=ϕϕ1=e.
Since Aut(G) satisfies all the group axioms, it forms a group under , as needed.

alenahelenash

alenahelenash

Expert

2022-01-24Added 366 answers

Solution: a) Let G be a group By definition Aut G={f:GG|f is an isomorphism of G} Claim: Aut G is group with binary operation composition i.e. (Aut(G), 0) is group Let f, gAut(G) 1) Closure: We have to show that gof is an autimorphism i.e. a homomorphism that is bijective Since g and f are bijective, so gof is bigective moreover a, bG (gof)(ab)=g(f(ab)) =g(f(a)f(b)) =g(f(a))×g(f(b)) =(gof)(a)×(gof)(b) g×f is closed under composition in Aut (G) 2) Associetively: Let f, g, hAut(G) & aG To prove: (hog)of=ho(gof) It is obvious duc to associativity holds is G. 3) Existence of Identily: Clearly. IdG:GG defined a aa is an aotomorphism Hore, Id: means identify mapping fIdG=IdGf, fAut(G) IdG is a natural (i.e. identity) element 4) Existence of inverse: Suppose fAut(G) Clearly, f1f=IdG=ff1 Aut(G) is a group under composition i.e. (Aut(G), 0) is a group b) Consider δ3 be a group which is abelian but whose Automorhism is non-abelian group Aut(C2×C2)δ3 Explanation: In δ3 i.e. symmetric group of 3 elements Let a=(1,2,3) & b=(1,2) ab=(1,3) & ba(2,3) ab ba=(1,3,2) & ba ab=(1,2,3) Automorphism given by a and conjugation by b do not conjugate Aut(G) is non-abelian while G is abelian.

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