1) Let R be the set of all real numbers. For all a * b = e^(a+b), where e^(x) de

Question

1) Let R be the set of all real numbers. For all a * b = e^(a+b), where e^(x) denotes the natural exponential function at x. Is R with * a multiplicative group?

2) Let S be a nonempty set. Denote by Sym(s) the set of all functions f: S -> s such that f is bijective. For all f and g in sym(S), define fg = f o g, where o denotes composition of functions. Prove that sym(s) with this operation is a group, which is in general non-abelian.

3) Construct the Cayley tables for the groups Z/5, Gamma 5, U(Z/20) and U (Z/12).

Explanation / Answer

1) Associativity fails since :

(a*b)*c = e^(a+b)*c = e^(e^(a+b)+c) and a*(b*c) = e^(a+e^(b+c)) are different

So this is not a multiplicative group

2)

a) Suppose f,g are in sym(s) , then fg = fog is bijective, because the composition of two bijectives function is bijective.

b) You know that composition is associative, that is : (fog)oh = fo(goh) , so f(gh) = (fg)h also.

c) Since f is a bijection, you know that f o f^(-1) = Id, so ff^(-1) = Id

So sym(s) with this composition operation is indeed a group. He is non-abelian in general since fog != gof

For example Take f = 1 , and g = 2 constant functions ,   fog = 1 and gof = 2 ....

3 ) Is a little longer, please wait :)

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