Can there really be an associative, but non-commutative binary operation with a identity and inverse?
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2Welcome 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. – Hrodelbert Jun 03 '15 at 09:14
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1Such a structure is called a (non-commutative) group. – Najib Idrissi Jun 03 '15 at 09:36
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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 $n$.
5xum
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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.
Hrodelbert
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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.
Emilio Novati
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