§2 Groups (Proofs)

There is only one identity.Proof: Suppose e and e' are identities. e∙e' = e' = e∎

Proposition 2.1:

Proposition 2.2:

Every element of G has only one inverse.Proof: Suppose b,c are inverses of a. ⟹ a∙b = a∙c = e      b∙a = c∙a = e b = b*e = b(ac) = (ba)c = e∙c = c ∴ b=c ⟹ there is only one inverse.∎

Proposition 2.3:

∀ a,b∈G, \((a*b)^{-1}\) = \(a^{-1}\)*\(b^{-1}\)Proof:Show (\(a^{-1}\)*\(b^{-1}\))*(a*b) = e (\(a^{-1}\)*\(b^{-1}\))*(a*b) = \(b^{-1}\)*(\(a^{-1}\)*a)*b                                                 = \(b^{-1}\)*e*b                                                 = e(a*b)*((\(b^{-1}\)*\(a^{-1}\)) = e automatically follows∎

Proposition 2.4:

For every a∈G, \((a^{-1})^{-1}\)= aProof:We have \((a^{-1})^{-1}\)*\(a^{-1}\) = e⟹ \((a^{-1})^{-1}\)*\(a^{-1}\)*a = e*a⟹ \((a^{-1})^{-1}\) = a∎

Proposition 2.5: Cancellation Law

Let (G,*�) be a group. For any a,b,c∈G, if a*b = a*c then b = c. Similarly if b*�a=c*a then b = c.Proof:Suppose a*b = a*c. Then \(a^{-1}\)*a*b = \(a^{-1}\)*a*c, hence b = c.Suppose b*a = c*a. Then b*a*\(a^{-1}\) = c*a*\(a^{-1}\), hence b = c.∎

Proposition 2.6:

Let (G,*�) be a group. For any a,b∈G, there is exactly one element x such that a*x = b and there is exactly one element x'� such that x'*a = b.Proof:We first prove the existence of x and x'�. Let x =  \(a^{-1}\)*b and x'� = b* \(a^{-1}\).We have a�*\(a^{-1}\)*�b = b and b*\(a^{-1}\)*�a = b. Let us now prove uniqueness of x. Suppose a*\(x_1\) = b and a*\(x_2\) = b for elements \(x_1\),\(x_2\)∈G. Thus a*\(x_1\) = a*\(x_2\) and by the cancellation law (Proposition 2.5) we have \(x_1\) = \(x_2\). The proof of uniqueness of x'� is similar.∎

Corollary 2.1:

iii) G → G, {\(a_1\),...,\(a_n\)}→ {\(ga_1\),...,\(ga_n\)} is one-to-one. Similarly, {\(a_1\),...,\(a_n\)}→ {\(ag_1\),...,\(ag_n\)} is one-to-one.Proof (iii) from notes:Onto? ∀ h∈g, there is a g such that gx=h.Yes, x = \(g^{-1}\)h, so the first map is onto.