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Can there really be an associative, but non-commutative binary operation with a identity and inverse?

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    $\begingroup$ Welcome to Math.SE! to make your question more useful to the community, it might be helpful why you seem to doubt that this is possible. $\endgroup$
    – Hrodelbert
    Jun 3 '15 at 9:14
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    $\begingroup$ Such a structure is called a (non-commutative) group. $\endgroup$ Jun 3 '15 at 9:36
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Yes, matrix multiplication: given two square matrices $A,B$ with non-zero determinant (so elements of $GL(n,\mathbb{R})$) of any size, we know that $AB$ does not have to be equal to $BA$, but $A(BC) = (AB)C$, the identity matrix serves as an identity and $A^{-1}$ is defined uniquely.

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Yes, and there are plenty of them:

  • For any set $X$, you can define a binary operation $\circ$ on the set of mappings $f:X\to X$ as $(f\circ g)(x) = f(g(x))$ (composition). This operation, in general is not commutative, but it is associative and has an inverse.
  • On the set of all invertible matrices of size $n\times n$, the standard binary operation of multiplying matrices is not commutative. It has an inverse, and is associative.
  • For any $n$, the set of all permutations of $n$ elements has a non-commutative (if $n>2$) associative operation with an inverse.

In fact, most groups studied in group theory are non-abelian (meaning their operation is not commutative). For any $n$, the number of finite groups of size at most $n$ is much larger than the number of finite Abelian groups of size at most $n$.

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Yes. The Quaternions are an extension of real and complex numbers that is a skew field. This means that there are two operations (a sum and a product) that have the same properties as the usual operations on real (or complex) numbers and also division (by non zero elements) is always possible, but the product is not commutative.

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