Question 935176: fa(x)=(a^-1).xa how to show fa is homomorphism..
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! A homomorphism is a map between two groups which respects the group structure.
Let G be a group and let a be a fixed element of G. Define a function f:G-->G by fa(x)=a^(-1)xa for all x in G.
Let a' = a^(-1).
a) fa(xy) = a'xya [definition]
= a'x(aa')ya [existence of inverse; identity]
= (a'xa)(a'ya) [associativity]
= fa(x)fa(y) [definition]
b) fa(x) = fa(y) implies a'xa = a'ya. Right multiplying by a' yields a'x = a'y, and Left multiplying by a yields x = y. Consequently fa(x) = fa(y) --> x = y, the definition of one-one.
c) Let g in G be given. fa(aga') = a'(aga')a = (a'a)g(a'a) = g. By closure, aga' is in G. Consequently, for each g in G, there is an x in G (specifically, x = aga') so that f(x) = g -- the definition of onto.
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