§2 Groups (Proofs)

Proposition 2.1:

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

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, \((a^{-1})^{-1}\)= a
• Proof:
• 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.