We are tasked with proving that if $a$ and $b$ are elements of a monoid $M$, and they commute, meaning $a * b = b * a$, then $(a * b) * (a * b) = (a * a) * (b * b)$.

Proof:

We know the following:

• $a$ and $b$ are elements of a monoid $M$, so there exists an associative binary operation $*$ (denoted as multiplication for simplicity).
• The identity element of the monoid is denoted as $e$, and the operation is associative: $x * (y * z) = (x * y) * z$ for any $x, y, z \in M$.
• It is also given that $a$ and $b$ commute, i.e., $a * b = b * a$.

We need to show:

$(a * b) * (a * b) = (a * a) * (b * b)$

Left-hand side:

We start by expanding $(a * b) * (a * b)$ using the associativity of the monoid operation:

$(a * b) * (a * b) = a * (b * (a * b))$

Since $a * b = b * a$ by assumption (commutativity), we can replace $b * a$ with $a * b$:

$a * (b * (a * b)) = a * ((b * a) * b) = a * ((a * b) * b)$

Now use the associativity of the operation again:

$a * ((a * b) * b) = (a * a) * (b * b)$

Thus, the left-hand side simplifies to $(a * a) * (b * b)$.

Conclusion:

Since both the left-hand side and right-hand side are equal, we have shown that:

$(a * b) * (a * b) = (a * a) * (b * b)$

The proof is complete.

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Would you like more details, or have any questions?

5 Related Questions:

1. What properties must a set and operation satisfy to form a monoid?
2. How does commutativity affect the structure of a monoid?
3. Can you prove a similar identity for semigroups, where an identity element is not required?
4. How would the proof change if $a$ and $b$ did not commute?
5. Can you extend this result to higher powers of $a * b$?

Tip:

Remember that commutativity simplifies many algebraic operations, but it's not guaranteed in all algebraic structures like monoids or groups. Always verify commutativity before using it!