Thread: Question about isomorphism 4?

Given a group G, let A(G) be the set of all isomorphism from G to itself. For every g∈G, define the map φg : G-->G by φg(x) = g^ -1xg
(a) Show that A(G) is a group under composition.
(b) Prove thatφg ∈A(G) for every g ∈G.
(c) Let φ: G-->A(G) be the map g-->φg. Verify that φ is a homomorphism.

Note φg, g is subscript to the φ (i do not know how to type)

Originally Posted by koukou8617

Given a group G, let A(G) be the set of all isomorphism from G to itself. For every g∈G, define the map φg : G-->G by φg(x) = g^ -1xg
(a) Show that A(G) is a group under composition.
(b) Prove thatφg ∈A(G) for every g ∈G.
(c) Let φ: G-->A(G) be the map g-->φg. Verify that φ is a homomorphism.

Note φg, g is subscript to the φ (i do not know how to type)

For (a), you need to verify the group axioms for A(G).
The identity element of A(G) is the trivial automorphism sending each element to itself and the inverse of an automorphism $\displaystyle \phi$ is its inverse $\displaystyle \phi^{-1}$ as a set map.

For (b) and (c), to show that $\displaystyle \phi_g$ is an automorphism, you need to show first that $\displaystyle \phi_g$ is well-defined. Then, you need to show that $\displaystyle \phi_g$ is homomorphism such that $\displaystyle \phi_g(xy)=\phi_g(x) \phi_g(y)$. To show injectivity of $\displaystyle \phi_g$, you need to show that if $\displaystyle \phi_g(x)=\phi_g(y)$, then x=y, where $\displaystyle x,y \in G$. To show surjectivity, it suffice to show that $\displaystyle x=\phi_g(gxg^{-1})$, where $\displaystyle x, g \in G$.

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This is some additional stuff.
Those maps in (b) is called an inner automorphism of G. If $\displaystyle \sigma \in Aut(G)$ and $\displaystyle \phi_g \in Inn(G)$, then $\displaystyle \sigma \phi_g \sigma^{-1} = \phi_{\sigma(g)}$, which shows $\displaystyle Inn(G) \triangleleft Aut(G) $. Additionally, outer automorphism can be defined as $\displaystyle Out(G)=Aut(G)/Inn(G) $.