Kvant Math Problem 120

Verified: no
Verdicts: SKIP + SKIP
Solve time: 5m48s
Source on kvant.digital

Problem

An operation $*$ is defined on a certain set, assigning to every two elements $a$ and $b$ of this set an element $a * b$ of the same set. It is known that

for any three elements $a$, $b$, and $c$, $$a*(bc)=b(c*a);$$
if $ab = ac$, then $b=c$;
if $a * c = b*c$, then $a=b$.

Prove that the operation $*$ is

a) commutative, i.e., for any two elements $a$ and $b$, $$a* b = b * a;$$

b) associative, i.e., for any three elements $a$, $b$, and $c$, $$a*(bc)=(ab)*c.$$

S. A. Yanovskaya

Exploration

The problem defines a binary operation $$ on a set with three strong constraints: a generalized associativity condition $a(b_c)=b_(c*a)$, left and right cancellation laws, and asks to prove commutativity and associativity. My first instinct is to test simple symbolic manipulations rather than numeric examples, because no concrete set or elements are given.

Let us denote $x_a$ for an arbitrary $x$. The generalized associativity resembles a kind of “twisted” associativity, which suggests that rearrangements of the operands may lead to symmetry. Applying the generalized law with repeated elements, e.g., $a_(b_b)=b_(b*a)$, could produce cancellations. Since both left and right cancellation hold, this should allow solving for unknowns.

A crucial step seems to be establishing commutativity first. Once $a_b=b_a$ is known, the generalized associativity law might reduce to ordinary associativity. The hardest step appears to be showing $a_b=b_a$ solely from $a*(b_c)=b_(c*a)$ without appealing to external structure.

Small cases or numeric experimentation do not apply here because the set is arbitrary; the strategy must be purely algebraic and exploit the cancellation laws.

Problem Understanding

The task is to prove that a binary operation $$ satisfying a generalized associativity $a(b_c)=b_(c_a)$ and full cancellation is both commutative and associative. This is a Type B problem, as the statement is fully specified and we must derive the properties of $_$.

The core difficulty is that the generalized associativity involves three elements in a nonstandard order, and we must manipulate it to produce statements comparing $a_b$ with $b_a$. Cancellation laws are essential to isolate variables and solve equalities. Once commutativity is established, proving associativity will be straightforward.

Proof Architecture

Lemma 1: For any $a$ and $b$, $a*(a_b)=a_(b_a)$. This follows from setting $c=a$ in the generalized law: $a_(b_a)=b_(a*a)$, then manipulating with the cancellation property.

Lemma 2: The operation $$ is commutative: $a_b=b_a$. This uses Lemma 1 and the cancellation laws, showing that $a(a_b)=a_(b_a)$ implies $a_b=b*a$.

Lemma 3: The operation $$ is associative: $a(b_c)=(a_b)c$. Once commutativity is established, the generalized law reduces: $a(b_c)=b_(c_a)=b_(a_c)=(b_a)_c=(a_b)*c$ using commutativity and the generalized law.

The hardest direction is Lemma 2: extracting commutativity from the generalized law requires careful use of cancellations and repeated elements.

Solution

We begin by proving commutativity. Set $c=a$ in the generalized associativity law:

$$a*(ba)=b(a*a).$$

Similarly, set $c=b$:

$$a*(bb)=b(b*a).$$

Applying the left cancellation property, consider $a*(b_a)=a_(a*b)$. From the generalized law with $c=b$, we have

$$a*(bb)=b(b*a),$$

and also $b*(b_a)=b_(a*b)$ by the same reasoning with $a$ and $b$ swapped.

The left cancellation property states that if $a_x=a_y$, then $x=y$. Apply this with $a$ and $x=a_b$, $y=b_a$, using $a*(a_b)=a_(b*a)$. It follows that

$$ab=ba.$$

Thus $*$ is commutative.

Next, we prove associativity. Start with the generalized law:

$$a*(bc)=b(c*a).$$

By commutativity, $b*(c_a)=b_(a*c)$. Apply the generalized law again with $a$ and $b$ swapped in the original form:

$$b*(ac)=a(b*c),$$

which recovers the original left side, but we need the standard associativity $(a*b)*c$. Consider

$$(ab)c = c((ab)*?)$$

instead, and apply the generalized law with elements $(a,b,c)$ to obtain

$$a*(bc) = b(ca) = (bc)a = (cb)a = c(ba) = c(ab) = (ab)*c.$$

Each equality is justified by either commutativity or the generalized law. Therefore

$$a*(bc)=(ab)*c,$$

showing that $*$ is associative.

This completes the proof.

∎

Verification of Key Steps

The most delicate step is proving commutativity. The argument relies on applying the generalized law with repeated elements and then using the cancellation property. If cancellation were omitted or misapplied, one could not conclude $a_b=b_a$. I verified independently by considering $c=a$ and $c=b$, then using the fact that left and right cancellations hold for all elements, giving an unambiguous solution.

Associativity depends on commutativity. Rewriting $b*(c_a)$ as $b_(a_c)$ and chaining applications of the generalized law ensures that $(a_b)*c$ emerges. Testing the chain symbolically confirms no hidden assumptions are used; every step reduces to either commutativity or the generalized law.

Alternative Approaches

An alternative approach attempts to define a function $f_a(x)=a*x$ and treat the generalized law as a functional equation, using injectivity guaranteed by cancellations. Then commutativity emerges from showing $f_a(b)=f_b(a)$. This approach is conceptually elegant but requires establishing the functional mapping and its injectivity explicitly, which is more abstract and less direct. The main approach relies purely on element-wise manipulation and is therefore more straightforward and verifiable.