bianary operation

Question

Let A = Q\{1}. Consider the binary operation "*" define of A by

a:b=a+b—ab, for every a,b € A.

How

i, Is "°" commutative?

iii, Show that A is a group.

i, Let @:G + G defined by O(x) = e, for all x € G, where e is the identity

element, Show that @ is a homomorphism,

i ne={[F | ren}, s:G +H define by f(r) = [> ‘| for al [7 ‘| €

G. Show that f is an isomorphism. Where (G, +) and (H, +) are groups in

usual notation,

Answer 1

1a) I understand Q to be the set of rational numbers (a/b where a and b are integers) and A, in this case, to be this set excluding the number 1, that is, a/b≠1 and a≠b. (b≠0 is another condition.)

(i) The operation is commutative because a and b can be interchanged:

a:b=b:a=a+b-ab=b+a-ba. This is because addition and multiplication are themselves commutative.

(iii) A={a/b | a,b∈Z, a≠b}. So A={..., -¼, -⅓, -½, ½, ⅓, ¼, ... 2, 3, 4, ..., -2, -3, -4, ..., ⅔, ⅖, ..., 3/2, ¾, 5/2, ...} for example. A is a group. Note that, for example, 2=2/1=4/2=6/3=..., ½=2/4=3/6=... so fractions have multiple representation a/b=c/d, but a,b,c,d∈Z for some a, b, c, d.

1b)

(i)  Homomorphism requires f(a*b)=f(a)*f(b) for a function f. Given φ:G→G, φ(x)=e ∀ x∈G, to show φ(x*y)=φ(x)*φ(y). φ(x*y)=e; φ(x)=e, φ(y)=e, and e*e=e by definition of identity. So φ(x*y)=φ(x)*φ(y), satisfying homomorphism.

Isomorphism is a special case of homomorphism in that it's also bijective (unique mapping).

More to follow in due course...