 Algotssleeddynf

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 alkaholikd9

Expert

Step 1
Let GG be a group. By definition

Given two automorphisms , we can consider the composition $g×f$. We claim that  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,
$\left(g×f\right)\left(ab\right)=g\left(f\left(ab\right)\right)=g\left(f\left(a\right)f\left(b\right)\right)=g\left(f\left(a\right)\right)g\left(f\left(b\right)\right)=\left(g×f\right)\left(a\right)\left(g×f\right)\left(b\right),$
for all . Hence $g×f$ is a group homomorphism.
Second, we must demonstrate that $×$ is associative, i.e. $\left(h×g\right)×f=h×\left(g×f\right)$. 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 $×$. Clearly $IdG:G\to G:a\to a$ is an automorphism. Since $f×IdG=IdG×f$ for all  is the neutral element.
Last but not least, we must ensure that each $f\in Aut\left(G\right)$ has an inverse for $×$. Consider the inverse function $f-1$. Clearly $f-1×f=IdG=f×f-1$. 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 . By definition there exist a unique  such that $f\left(x\right)=a$ 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

Step 1
Let Aut(G) be the set of all automorphisms $\varphi :G\to G$. 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 , then it is not too tough to show that $\psi \circ \varphi \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\varphi \circ \psi$ 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,
$\psi \circ \varphi \left(ab\right)=\pi \left(\varphi \left(ab\right)\right)=\psi \left(\varphi \left(a\right)\varphi \left(b\right)\right)$,
because $\varphi$ is an isomorphism. Since $\psi$ is also an isomorphism,
$\psi \left(\varphi \left(a\right)\varphi \left(b\right)\right)=\psi \circ \varphi \left(a\right)\psi \circ \varphi \left(b\right)$,
so the composition $\psi \circ \varphi$ preserves products. Thus, $\psi \circ \varphi$ is an isomorphism if $\psi$ and $\varphi$ are.
2) Is the set associative? Yes! All you need to do is show that, for any three isomorphisms and
.
To do that, just show that for each x in
$G,\varphi \circ \left(\psi \circ \xi \right)\left(x\right)=\left(\varphi \circ \psi \right)\circ \xi \left(x\right)=\varphi \left(\psi \left(\xi \left(x\right)\right)\right)$.
It's just pushing around definitions.
3) Does the set contain an identity element? Yes! Let the identity automorphism $e:G\to G$ be the map $e\left(x\right)=x$. Clearly, $e\circ \varphi =\varphi \circ e=\varphi$.
4) Does each element of the set have an inverse under $\circ$? Yes! Since each isomorphism $\varphi :G\to G$ is bijective, there is a well-defined inverse map ${\varphi }^{-1}:G\to G$. 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
${\varphi }^{-1}\circ \varphi =\varphi \circ {\varphi }^{-1}=e$.
Since Aut(G) satisfies all the group axioms, it forms a group under $\circ$, as needed. alenahelenash

Expert

Solution: a) Let G be a group By definition Aut Claim: Aut G is group with binary operation composition i.e. is group Let 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 $\left(gof\right)\left(ab\right)=g\left(f\left(ab\right)\right)$ $=g\left(f\left(a\right)f\left(b\right)\right)$ $=g\left(f\left(a\right)\right)×g\left(f\left(b\right)\right)$ $=\left(gof\right)\left(a\right)×\left(gof\right)\left(b\right)$ $\therefore g×f$ is closed under composition in Aut (G) 2) Associetively: Let To prove: $\left(hog\right)of=ho\left(gof\right)$ 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 $\therefore I{d}_{G}$ is a natural (i.e. identity) element 4) Existence of inverse: Suppose $f\in Aut\left(G\right)$ Clearly, ${f}^{-1}\circ f=I{d}_{G}=f\circ {f}^{-1}$ $\therefore Aut\left(G\right)$ is a group under composition i.e. is a group b) Consider ${\delta }_{3}$ be a group which is abelian but whose Automorhism is non-abelian group $Aut\left({C}_{2}×{C}_{2}\right)\cong {\delta }_{3}$ Explanation: In ${\delta }_{3}$ i.e. symmetric group of 3 elements Let $\therefore$ Automorphism given by a and conjugation by b do not conjugate $⇒Aut\left(G\right)$ is non-abelian while G is abelian.