群中二元运算的单位元是否唯一？实数集上a*b=b的运算是否构成群？

Hey folks, let's break down these two group theory questions one by one:

Absolutely yes! Let me walk through the quick proof to confirm this:

Suppose there are two identity elements ( e_1 ) and ( e_2 ) in the group. By the core definition of a group identity:

• ( e_1 * e_2 = e_1 ) (since ( e_2 ) acts as an identity, it leaves ( e_1 ) unchanged when operated on the right)
• ( e_1 * e_2 = e_2 ) (since ( e_1 ) acts as an identity, it leaves ( e_2 ) unchanged when operated on the left)

From these two equalities, we directly get ( e_1 = e_2 ). So every group has exactly one unique identity element—no exceptions.

Let's verify against the three group axioms your textbook lays out:

1. Associativity: Does ( (ab)c = a(bc) ) hold for all real numbers ( a, b, c )?

   • Left side: ( (a*b)c = bc = c )
   • Right side: ( a*(bc) = ac = c )
     Yep, associativity is satisfied here.

2. Existence of a global identity element: This is where we hit a critical roadblock. The textbook's definition requires a single identity element ( e ) that works for every element ( a ) in the set—meaning ( ea = a ) AND ( ae = a ) must both be true for all ( a \in \mathbb{R} ).

   • Let's test the second condition first: ( ae = a ). But per the operation rule, ( ae = e ), so this would require ( e = a ) for every real number ( a ). That's impossible—there's no single real number that equals every other real number.
   • The note you mentioned about "each element's identity being itself" doesn't fit the group definition. Groups demand one universal identity, not a unique "identity" per element.

3. Existence of inverses: Since we don't have a valid global identity element, we can't even define inverses (inverses rely on the identity to satisfy ( a*a^{-1} = e ) and ( a^{-1}*a = e )). This axiom is automatically failed.

Final verdict: This set and operation do not form a group because they fail the "existence of a global identity element" requirement.

内容的提问来源于stack exchange，提问作者Tandeitnik