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

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

Expert

2022-01-13Added 37 answers

Step 1

Let GG be a group. By definition

$Aut\left(G\right)=\{f:G\to G\mid f\text{}\text{is an isomorphism of}\text{}G\}.$

Given two automorphisms $f,\text{}g\in Aut\left(G\right)$, we can consider the composition $g\times f$. We claim that $Aut\left(G\right),\text{}\times$ is a group. We have to check all axioms.

First of all we need to show that $g\times f$ is again an automorphism, i.e. a homomorphism that is bijective. Now since g and f are bijective, $g\times f$ is bijective. Moreover,

$(g\times f)\left(ab\right)=g\left(f\left(ab\right)\right)=g\left(f\left(a\right)f\left(b\right)\right)=g\left(f\left(a\right)\right)g\left(f\left(b\right)\right)=(g\times f)\left(a\right)(g\times f)\left(b\right),$

for all $a,\text{}b\in G$. Hence $g\times f$ is a group homomorphism.

Second, we must demonstrate that $\times$ is associative, i.e. $(h\times g)\times f=h\times (g\times f)$. Just evaluate both morphisms at $a\in G$ and see that both expressions coincide due to the associativity of GG.

Thirdly, we must confirm that a neutral component exists for $\times$. Clearly $IdG:G\to G:a\to a$ is an automorphism. Since $f\times IdG=IdG\times f$ for all $f\in Aut\left(G\right),\text{}IdG$ is the neutral element.

Last but not least, we must ensure that each $f\in Aut\left(G\right)$ has an inverse for $\times$. Consider the inverse function $f-1$. Clearly $f-1\times f=IdG=f\times f-1$. So it remains to show that $f-1$ 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 $f:G\to G$ is a group isomorphism. We need to show that $f-1$ is a group morphism. Let $a,\text{}b\in G$. By definition there exist a unique $x,\text{}y\in G$ such that $f(x)=a$ and $f\left(y\right)=b$. Hence

$f-1\left(ab\right)=f-1\left(f\left(x\right)f\left(y\right)\right)=f-1\left(f\left(xy\right)\right)=xy$.

Similarly

$f-1\left(a\right)f-1\left(b\right)=f-1\left(f\left(x\right)\right)f-1\left(f\left(y\right)\right)=xy$.

Therefore, $f-1\left(ab\right)=f-1\left(a\right)f-1\left(b\right)$

Stella Calderon

Expert

2022-01-14Added 35 answers

Step 1

Let Aut(G) be the set of all automorphisms

1) Is the set is closed under composition? Yes! If you are given isomorphisms

because

so the composition

2) Is the set associative? Yes! All you need to do is show that, for any three isomorphisms

To do that, just show that for each x in

It's just pushing around definitions.

3) Does the set contain an identity element? Yes! Let the identity automorphism

4) Does each element of the set have an inverse under

Since Aut(G) satisfies all the group axioms, it forms a group under

alenahelenash

Expert

2022-01-24Added 366 answers

Solution:
a) Let G be a group
By definition
Aut $G=\{f:G\to G|f\text{}\text{is an isomorphism of G}\}$
Claim: Aut G is group with binary operation composition i.e. $(Aut(G),\text{}0)$ is group
Let $f,\text{}g\in Aut(G)$
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 $\mathrm{\forall}a,\text{}b\in G$
$(gof)(ab)=g(f(ab))$
$=g(f(a)f(b))$
$=g(f(a))\times g(f(b))$
$=(gof)(a)\times (gof)(b)$
$\therefore g\times f$ is closed under composition in Aut (G)
2) Associetively:
Let $f,\text{}g,\text{}h\in Aut(G)\text{}\mathrm{}\text{}a\in G$
To prove: $(hog)of=ho(gof)$
It is obvious duc to associativity holds is G.
3) Existence of Identily:
Clearly. $\mathrm{\exists}I{d}_{G}:G\to G$ defined a $a\to a$ is an aotomorphism
Hore, Id: means identify mapping
$\therefore f\circ I{d}_{G}=I{d}_{G}\circ f,\text{}\mathrm{\forall}f\in Aut(G)$
$\therefore I{d}_{G}$ is a natural (i.e. identity) element
4) Existence of inverse:
Suppose $f\in Aut(G)$
Clearly, ${f}^{-1}\circ f=I{d}_{G}=f\circ {f}^{-1}$
$\therefore Aut(G)$ is a group under composition
i.e. $(Aut(G),\text{}0)$ is a group
b) Consider ${\delta}_{3}$ be a group which is abelian but whose
Automorhism is non-abelian group
$Aut({C}_{2}\times {C}_{2})\cong {\delta}_{3}$
Explanation: In ${\delta}_{3}$ i.e. symmetric group of 3 elements
Let $a=(1,2,3)\text{}\mathrm{}\text{}b=(1,2)$
$\therefore ab=(1,3)\text{}\mathrm{}\text{}ba(2,3)$
$\Rightarrow ab\text{}ba=(1,3,2)\text{}\mathrm{}\text{}ba\text{}ab=(1,2,3)$
$\therefore $ Automorphism given by a and conjugation by b do not conjugate
$\Rightarrow Aut(G)$ is non-abelian while G is abelian.

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