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Old 07-21-2017, 05:28 PM
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lpetrich lpetrich is offline
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Default Re: Math trivia

Now for something different. Abstract algebra. In particular, group theory.

I start with a binary operation * over some set of entities S, an operation that is closed over it: S*S = S.

I add associativity and I get a semigroup: (a*b)*c = a*(b*c). Commutativity is optional.

I add an identity element and I get a monoid: element e such that a*e = e*a = a.

It is easy to prove that an identity element is unique -- a monoid cannot contain more than one distinct identity.

I add inverses and I get a group: for all a in S, an inverse inv(a) satisfying a*inv(a) = inv(a)*a = e.

Commutative groups are usually called abelian groups.

One can easily make new groups from products of existing ones:
(a1,a2) * (b1,b2) = (a1*b1, a2*b2)

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The integers under addition form a group, as do the rational numbers under addition, the real numbers under addition, the nonzero rational numbers under multiplication, the nonzero real numbers under multiplication, etc.

What are the identities and the inverses in those groups?

The group of integers under addition is called Z, and that of integers modulo n Z[sub]n[/sup] or Z/nZ or Z(n). One can picture the integers with a numberline, and for the modulo case, one can wrap part of that line around so that its ends join, making a circle.

The groups Z(n) are called cyclic groups, for obvious reasons. An element a generates it if every element in it is some power of a. That is, every element can be found from some a*a*...*a.

Every finite abelian (commutative) group is a product of cyclic groups, and cyclic groups themselves may be decomposed further with the help of the Chinese Remainder Theorem:

For n1 and n2 relatively prime (coprime): Z(n1*n2) = Z(n1)*Z(n2).

That means that groups Z(pn) for prime p cannot be reduced further in this fashion, and that every finite abelian group is thus the product of such groups.

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Let us now consider subgroups. A group G has a group H as a subgroup if H has the same operation and if H's elements are all in G's elements. Now consider the order or number of elements of a group. Lagrange's theorem states that (order of H) evenly divides (order of G).

This constrains what subgroups a group can have. Every group has two trivial subgroups, itself and the identity group {e} or Z(1). The group Z(n) thus only has subgroups Z(m) where m evenly divides n. For n a prime, Z(n) only has the two trivial subgroups.

The order of an element a in a group is the number of a's n in an = a*a*...*a that give e such that no smaller number does so. It's easy to see that a generates cyclic group Z(n), and by Lagrange's theorem, n must divide the order of the group.

If a generates Z(n), then some power of a must also generate a cyclic group, either Z(n) itself or some Z(m) where m evenly divides n.

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One can prove Lagrange's theorem by constructing cosets of a subgroup H of G:

Left cosets: a1*H, a2*H, a3*H, ...
Right cosets: H*a1, H*a2, H*a3, ...
Each side's cosets are disjoint and they all cover G.

If (left cosets) = (right cosets), then H is called a normal subgroup of G. In this case, one can define a coset multiplication law: (a1*H) * (a2*H) = (a3*H), and with the cosets, it forms a group. This is the quotient group or factor group of G and H: G/H.

I'll find these groups in a simple case: Z(4). Its elements are {e, a, a2, a3}. It has subgroup Z(2): {e, a2}. Its cosets are (e) = {e, a2} and (a) = {a, a3}, with multiplication law
(e) * (e) = (e)
(e) * (a) = (a)
(a) * (e) = (a)
(a) * (a) = (e)
Thus being Z(2). So, Z(4) / Z(2) = Z(2).

Likewise, the integer group Z has a normal subgroup, the even numbers: 2*Z. It has cosets 2*Z and the odd numbers 2*Z+1. From
even + even = even
even + odd = odd
odd + even = odd
odd + odd = even,
it is evident that Z / (2*Z) = Z(2).

In fact, every subgroup with only two cosets is a normal subgroup.

I think I'll stop here.
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Thanks, from:
But (08-16-2017), ceptimus (07-21-2017), Ensign Steve (07-21-2017)
 

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