Answer sir questions.
1. (a) Define inner automorphism of a group. Prove that the set I (G) of all inner
automorphism of a group G of allautomorphisms of G.
Answers
Answer:
I have shown that Inn(G) is non-empty and that it is closed under composition of functions. I am now trying to show that the inverses of inner automorphisms are also inner automorphisms.
This is what I have so far in this regard:
f:G→G is an inner automorphism with f(x)=c−1xc for some fixed c∈G. If a∈G then since f is an isomorphism, it is also a bijection and so there is some x,y∈G where f(x)=a and f(y)=c. It follows x=f−1(a) and y=f−1(c). Hence
f−1(a)=f−1(f(x))=f−1(c−1xc)=[f−1(c)]−1f−1(x)f−1(c)=y−1f−1(x)y.
At this point I want f−1(x)=a but then x=f(a) which seems suspicious as then x=c−1ac so then f(x)=f(c−1ac)=(c2)−1ac2=a so (c2)−1a=a(c2)−1 but we aren't assuming G is abelian
Explanation:
I have shown that Inn(G) is non-empty and that it is closed under composition of functions. I am now trying to show that the inverses of inner automorphisms are also inner automorphisms.
This is what I have so far in this regard:
f:G→G is an inner automorphism with f(x)=c−1xc for some fixed c∈G. If a∈G then since f is an isomorphism, it is also a bijection and so there is some x,y∈G where f(x)=a and f(y)=c. It follows x=f−1(a) and y=f−1(c). Hence
f−1(a)=f−1(f(x))=f−1(c−1xc)=[f−1(c)]−1f−1(x)f−1(c)=y−1f−1(x)y.
At this point I want f−1(x)=a but then x=f(a) which seems suspicious as then x=c−1ac so then f(x)=f(c−1ac)=(c2)−1ac2=a so (c2)−1a=a(c2)−1 but we aren't assuming G is abelian