Ciro Santilli @cirosantilli 34

 Incoming links: Direct product of groups

Abelian group  Updated 2024-12-15  +Created 1970-01-01

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Easily classified as the direct product of cyclic groups of prime order.

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Connected components of the orthogonal group  Updated 2024-12-15  +Created 1970-01-01

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The orthogonal group has 2 connected components:

one with determinant +1, which is itself a subgroup known as the special orthogonal group. These are pure rotations without a reflection.
the other with determinant -1. This is not a subgroup as it does not contain the origin. It represents rotations with a reflection.

It is instructive to visualize how the

\pm 1

looks like in

S O (3)

you take the first basis vector and move it to any other. You have therefore two angular parameters.
you take the second one, and move it to be orthogonal to the first new vector. (you can choose a circle around the first new vector, and so you have another angular parameter.
at last, for the last one, there are only two choices that are orthogonal to both previous ones, one in each direction. It is this directio, relative to the others, that determines the "has a reflection or not" thing

As a result it is isomorphic to the direct product of the special orthogonal group by the cyclic group of order 2:

O (n) ≅ S O (n) \times C_{2}

(1)

A low dimensional example:

O (1) ≅ S O (2) \times C_{2}

(2)

because you can only do two things: to flip or not to flip the line around zero.

Note that having the determinant plus or minus 1 is not a definition: there are non-orthogonal groups with determinant plus or minus 1. This is just a property. E.g.:

M = [2132]

(3)

has determinant 1, but:

M^{T} M = [58811]

(4)

M

is not orthogonal.

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Direct product  Updated 2024-12-15  +Created 1970-01-01

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A Cartesian product that carries over some extra structure of the input groups.

E.g. the direct product of groups carries over group structure on both sides.

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Group extension problem  Updated 2024-12-15  +Created 1970-01-01

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Besides the understandable Wikipedia definition, Video "Simple Groups - Abstract Algebra by Socratica (2018)" gives an understandable one:

Given a finite group $F$ and a simple group $S$ , find all groups $G$ such that $N$ is a normal subgroup of $G$ and $G / N = S$ .

We don't really know how to make up larger groups from smaller simple groups, which would complete the classification of finite groups:

math.stackexchange.com/questions/25315/how-is-a-group-made-up-of-simple-groups

In particular, this is hard because you can't just take the direct product of groups to retrieve the original group: Section "Relationship between the quotient group and direct products".

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Relationship between the quotient group and direct products  Updated 2024-12-15  +Created 1970-01-01

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Although quotients look a bit real number division, there are some important differences with the "group analog of multiplication" of direct product of groups.

If a group is isomorphic to the direct product of groups, we can take a quotient of the product to retrieve one of the groups, which is somewhat analogous to division: math.stackexchange.com/questions/723707/how-is-the-quotient-group-related-to-the-direct-product-group

The "converse" is not always true however: a group does not need to be isomorphic to the product of one of its normal subgroups and the associated quotient group. The wiki page provides an example:

Given G and a normal subgroup N, then G is a group extension of G/N by N. One could ask whether this extension is trivial or split; in other words, one could ask whether G is a direct product or semidirect product of N and G/N. This is a special case of the extension problem. An example where the extension is not split is as follows: Let $G = Z 4 = 0, 1, 2, 3$ , and $= 0, 2$ which is isomorphic to Z2. Then G/N is also isomorphic to Z2. But Z2 has only the trivial automorphism, so the only semi-direct product of N and G/N is the direct product. Since Z4 is different from Z2 × Z2, we conclude that G is not a semi-direct product of N and G/N.

TODO find a less minimal but possibly more important example.

This is also semi mentioned at: math.stackexchange.com/questions/1596500/when-is-a-group-isomorphic-to-the-product-of-normal-subgroup-and-quotient-group

I think this might be equivalent to why the group extension problem is hard. If this relation were true, then taking the direct product would be the only way to make larger groups from normal subgroups/quotients. But it's not.

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Semidirect product  Updated 2024-12-15  +Created 1970-01-01

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As per en.wikipedia.org/w/index.php?title=Semidirect_product&oldid=1040813965#Properties, unlike the Direct product, the semidirect product of two goups is neither unique, nor does it always exist, and there is no known algorithmic way way to tell if one exists or not.

This is because reaching the "output" of the semidirect produt of two groups requires extra non-obvious information that might not exist. This is because the semi-direct product is based on the product of group subsets. So you start with two small and completely independent groups, and it is not obvious how to join them up, i.e. how to define the group operation of the product group that is compatible with that of the two smaller input groups. Contrast this with the Direct product, where the composition is simple: just use the group operation of each group on either side.

Product of group subsets

So in other words, it is not a function like the Direct product. The semidiret product is therefore more like a property of three groups.

The semidirect product is more general than the direct product of groups when thinking about the group extension problem, because with the direct product of groups, both subgroups of the larger group are necessarily also normal (trivial projection group homomorphism on either side), while for the semidirect product, only one of them does.

Conversely, en.wikipedia.org/w/index.php?title=Semidirect_product&oldid=1040813965 explains that if

G = N ⋊ H

, and besides the implied requirement that N is normal, H is also normal, then

G = N \times H

Smallest example:

D_{6} = C_{3} ⋊ C_{2}

where

D

is a dihedral group and

C

are cyclic groups.

C_{3}

(the rotation) is a normal subgroup of

D_{6}

, but

C_{2}

(the flip) is not.

Note that with the Direct product instead we get

C_{6}

and not

D_{6}

, i.e.

C_{3} \times C_{2} = C_{6}

as per the direct product of two cyclic groups of coprime order is another cyclic group.

TODO:

why does one of the groups have to be normal in the definition?
what is the smallest example of a non-simple group that is neither a direct nor a semi-direct product of any two other groups?

Bibliography: math.stackexchange.com/questions/1726939/is-this-intuition-for-the-semidirect-product-of-groups-correct

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