Basic question on binary operator that is both associative and commutative

Question

The example 74 in Sets and Groups by Green reads that "If the operation $\circ$ is both associative and commutative... it is easy to prove, for any $x,y\in A$ and any positive integer n, that $(x \circ y)^n=x^n\circ y^n$, for each side is simply a product of n factors equal to x with n factors equal to y."

If I choose x=4, y=3 and n=2 assuming $\circ$ is commutative and associative. In additive notation I'm getting $(7^2)\neq16+9$.

What I'm missing here?

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$x_1\circ x_1\circ \ldots \circ x_n$ (4.3)

"Example 74. If the operation $\circ$ is both associative and commutative, then the factors in the product (4.3) can be permuted in any way without altering its value. In particular it is easy to prove, for any $x,y\in A$ and any positive integer n, that $(x \circ y)^n=x^n\circ y^n$, for each side is simply a product of n factors equal to x with n factors equal to y. But if $\circ$ is not commutative this fails."

Answer 1

First, $16 \cdot 9 = 144 = 12^2$.

Second, if you chose additive notation $\circ = +$, then $x^2$ corresponds to $x\circ x = x + x = 2 x$. You hence get $$4^2 \circ 3^2 = 8 \circ 6 = 14 = 7^2$$

Answer 2

If $\circ$ is the sum of integers, then the assertion$$(4\circ3)^2=4^2\circ3^2$$means$$(4+3)+(4+3)=(4+4)+(3+3),$$which is indeed true.

Answer 3

Well, the multiplicative case is fine, since one writes multiples as powers.

But its wrong in the additive case, since one uses multiples:

$n(x\circ y) = nx \circ ny.$

In your example $2(3+4) = 2\cdot 3 + 2\cdot 4$.