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For the curious, it’s not a “group” because the integers don’t have multiplicative inverses. If I have x=2, there is no integer that I can multiply that by to g
by xeonmc 1y ago
For the curious, it’s not a “group” because the integers don’t have multiplicative inverses. If I have x=2, there is no integer that I can multiply that by to get 1. Integers with addition on the other hand is a group, which is a monoid with the additional property that inverses are present.
Would one be correct to say that square matrices are an example of monoids, since they have an identity element and are associative, but might not necessarily have inverses if their determinant is zero?
- Jaxan 1y agoYes
- sd9 1y agoThe operation is important too. Square matrices over integers with matrix multiplication is (just) a monoid. Square matrices with addition are a monoid too (but also a group, because there is an additive inverse). Put it all together and it’s called a ring
- ndriscoll 1y agoAdditionally, if you restrict to invertible matrices with matrix multiplication, they are also a monoid (so they are a submonoid), and in fact a group, the general linear group. If you restrict again to matrices with determinant 1, it's again itself a group (a subgroup, and also a submonoid), the special linear group.