Directly modelled by group.

For continuous symmetries, see: Lie group.

You just map the value (1, 1) $C_{m}×C_{n}$ to the value 1 of $C_{mn}$, and it works out. E.g. for $C_{2}×C_{3}$, the group generated by of (1, 1) is:

```
0 = (0, 0)
1 = (1, 1)
2 = (0, 2)
3 = (1, 0)
4 = (0, 1)
5 = (1, 2)
6 = (0, 0) = 0
```

A concise to describe a specific permutation.

A permutation group can then be described in terms of the generating set of a group of specific elements given in cycle notation.

E.g. en.wikipedia.org/w/index.php?title=Mathieu_group&oldid=1034060469#Permutation_groups mentions that the Mathieu group $M_{1}1$ is generated by three elements:which feels quite compact for a simple group with 95040 elements, doesn't it!

- (0123456789a)
- (0b)(1a)(25)(37)(48)(69)
- (26a7)(3945)

Suppose we have a given permutation group that acts on a set of n elements.

If we pick k elements of the set, the stabilizer subgroup of those k elements is a subgroup of the given permutation group that keeps those elements unchanged.

Note that an analogous definition can be given for non-finite groups. Also note that the case for all finite groups is covered by the permutation definition since all groups are isomorphic to a subgroup of the symmetric group

TODO existence and uniqueness. Existence is obvious for the identity permutation, but proper subgroup likely does not exist in general.

Bibliography:

Group of all permutations.

Or in other words: symmetric groups are boring, because they are basically everything already!

Group of even permutations.

Note that odd permutations don't form a subgroup of the symmetric group like the even permutations do, because the composition of two odd permutations is an even permutation.

www.youtube.com/watch?v=U_618kB6P1Q GT18.2. A_n is Simple (n ge 5) by MathDoctorBob (2012)

Our notation: $D_{n}$, called "dihedral group of degree n", means the dihedral group of the regular polygon with $n$ sides, and therefore has order $2n$ (all rotations + flips), called the "dihedral group of order 2n".

17 of them.

All possible repetitive crystal structures!

219 of them.

$C_{2}×C_{2}$

As shown in Video 1. "Simple Groups - Abstract Algebra by Socratica (2018)", this can be split up into two steps:This split is sometimes called the "Jordan-Hölder program" in reference to the authors of the jordan-Holder Theorem.

Good lists to start playing with:

This dude has done well.

Ciro Santilli is very fond of this result: the beauty of mathematics.

How can so much complexity come out from so few rules?

How can the proof be so long (thousands of papers)?? Surprise!!

And to top if all off, the awesomely named monster group could have a relationship with string theory via the monstrous moonshine?

all science is either physics or stamp collecting comes to mind.

The classification contains:

- cyclic groups: infinitely many, one for each prime order. Non-prime orders are not simple. These are the only Abelian ones.
- alternating groups of order 4 or greater: infinitely many
- groups of Lie type: a contains several infinite families
- sporadic groups: 26 or 27 of them depending on definitions

In the classification of finite simple groups, groups of Lie type are a set of infinite families of simple lie groups. These are the other infinite families besides te cyclic groups and alternating groups.

A decent list at: en.wikipedia.org/wiki/List_of_finite_simple_groups, en.wikipedia.org/wiki/Group_of_Lie_type is just too unclear. The groups of Lie type can be subdivided into:

- Chevalley groups
- TODO the rest

The first in this family discovered were a subset of the Chevalley groups $A_{n}(q)$ by Galois: $PSL(2,p)$, so it might be a good first one to try and understand what it looks like.

TODO understand intuitively why they are called of Lie type. Their names $A_{n}$, $B_{n}$ seem to correspond to the members of the classification of simple Lie groups which are also named like that.

But they are of course related to Lie groups, and as suggested at Video "Yang-Mills 1 by David Metzler (2011)" part 2, the continuity actually simplifies things.

They are the finite projective special linear groups.

This was the first infinite family of simple groups discovered after the simple cyclic groups and alternating groups. The first case discovered was $PSL(2,p)$ by Galois. You should understand that one first.

Examples of exceptional objects.

Contains the first sporadic groups discovered by far: 11 and 12 in 1861, and 22, 23 and 24 in 1973. And therefore presumably the simplest! The next sporadic ones discovered were the Janko groups, only in 1965!

Each $M_{n}$ is a permutation group on $n$ elements. There isn't an obvious algorithmic relationship between $n$ and the actual group.

TODO initial motivation? Why did Mathieu care about k-transitive groups?

Their; k-transitive group properties seem to be the main characterization, according to Wikipedia:Looking at the classification of k-transitive groups we see that the Mathieu groups are the only families of 4 and 5 transitive groups other than symmetric groups and alternating groups. 3-transitive is not as nice, so let's just say it is the stabilizer of $M_{2}3$ and be done with it.

- 22 is 3-transitive but not 4-transitive.
- four of them (11, 12, 23 and 24) are the only sporadic 4-transitive groups as per the classification of 4-transitive groups (no known simpler proof as of 2021), which sounds like a reasonable characterization. Note that 12 and 25 are also 5 transitive.

TODO why do we care about this?

Note that if a group is k-transitive, then it is also k-1-transitive.

TODO this would give a better motivation for the Mathieu group

Might be a bit complex: math.stackexchange.com/questions/698327/classification-of-triply-transitive-finite-groups

en.wikipedia.org/w/index.php?title=Mathieu_group&oldid=1034060469#Multiply_transitive_groups is a nice characterization of 4 of the Mathieu groups.

Apparently only Mathieu group $M_{1}2$ and Mathieu group $M_{2}4$.

www.maths.qmul.ac.uk/~pjc/pps/pps9.pdf mentions:

The automorphism group of the extended Golay code is the 54-transitive Mathieu group $M_{24}$. This is one of only two finite 5-transitive groups other than symmetric and alternating groupsHmm, is that 54, or more likely 5 and 4?

scite.ai/reports/4-homogeneous-groups-EAKY21 quotes link.springer.com/article/10.1007%2FBF01111290 which suggests that is is also another one of the Mathieu groups, math.stackexchange.com/questions/698327/classification-of-triply-transitive-finite-groups#comment7650505_3721840 and en.wikipedia.org/wiki/Mathieu_group_M12 mentions M_12.

math.stackexchange.com/questions/700235/is-there-an-easy-proof-for-the-classification-of-6-transitive-finite-groups says there aren't any non-boring ones.

A master thesis reviewing its results: scholarworks.sjsu.edu/cgi/viewcontent.cgi?referer=https://www.google.com/&httpsredir=1&article=5051&context=etd_theses

TODO clickbait, or is it that good?

Uniqueness results for the composition series of a group.

Besides the understandable Wikipedia definition, Video 1. "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:

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".

Something analogous to a group isomorphism, but that preserves whatever properties the given algebraic object has. E.g. for a field, we also have to preserve multiplication in addition to addition.

Other common examples include isomorphisms of vector spaces and field. But since both of those two are much simpler than groups in classification, as they are both determined by number of elements/dimension alone, see:we tend to not talk about isomorphisms so much in those contexts.

Like isomorphism, but does not have to be one-to-one: multiple different inputs can have the same output.

The image is as for any function smaller or equal in size as the domain of course.

This brings us to the key intuition about group homomorphisms: they are a way to split out a larger group into smaller groups that retains a subset of the original structure.

As shown by the fundamental theorem on homomorphisms, each group homomorphism is fully characterized by a normal subgroup of the domain.

You select a generating set of a group, and then you name every node with them, and you specify:

- each node by a product of generators
- each edge by what happens when you apply a generator to each element

Not unique: different generating sets lead to different graphs, see e.g. two possible en.wikipedia.org/w/index.php?title=Cayley_graph&oldid=1028775401#Examples for the

How to build it: math.stackexchange.com/questions/3137319/how-in-general-does-one-construct-a-cycle-graph-for-a-group/3162746#3162746 good answer with ASCII art. You basically just pick each element, and repeatedly apply it, and remove any path that has a longer version.

Immediately gives the generating set of a group by looking at elements adjacent to the origin, and more generally the order of each element.

TODO uniqueness: can two different groups have the same cycle graph? It does not seem to tell us how every element interact with every other element, only with itself. This is in contrast with the Cayley graph, which more accurately describes group structure (but does not give the order of elements as directly), so feels like it won't be unique.

Take the element and apply it to itself. Then again. And so on.

In the case of a finite group, you have to eventually reach the identity element again sooner or later, giving you the order of an element of a group.

The continuous analogue for the cycle of a group are the one parameter subgroups. In the continuous case, you sometimes reach identity again and to around infinitely many times (which always happens in the finite case), but sometimes you don't.

The length of its cycle.

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 $G=N×H$.

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. $C_{3}×C_{2}=C_{6}$ 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?

Does not have to be isomorphic to a subgroup:

This is one of the reasons why the analogy between simple groups of finite groups and prime numbers is limited.

Quotient of a subgroup H of G by a normal subgroup of the subgroup H.

That normal subgroup does not have have to be a normal subgroup of G.

As an overkill example, the happy family are subquotients of the monster group, but the monster group is simple.

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 $G=Z4=0,1,2,3$, 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.

Only normal subgroups can be used to form quotient groups: their key definition is that they plus their cosets form a group.

One key intuition is that "a normal subgroup is the kernel" of a group homomorphism, and the normal subgroup plus cosets are isomorphic to the image of the isomorphism, which is what the fundamental theorem on homomorphisms says.

Therefore "there aren't that many group homomorphism", and a normal subgroup it is a concrete and natural way to uniquely represent that homomorphism.

The best way to think about the, is to always think first: what is the homomorphism? And then work out everything else from there.

Does not have any non-trivial normal subgroup.

And therefore, going back to our intuition that due to the fundamental theorem on homomorphisms there is one normal group per homomorphism, a simple group is one that has no non-trivial homomorphisms.

scholarworks.sjsu.edu/cgi/viewcontent.cgi?referer=https://www.google.com/&httpsredir=1&article=5051&context=etd_theses proves that the Mathieu group $M_{2}4$ is simple in just 200 pages. Nice.

A group with an extra operation called multiplication which satisfies:

- associative property
- distributes over addition (the default group operation)
- has an identity

Unlike addition, that multiplication does not need to satisfy:If those are also satisfied, then we have a field.

- commutative property. If this is satisfied, we can call it a commutative ring.
- existence of an inverse. If this is satisfied, we can call it a division ring.

The simplest example of a ring which is not a full fledged field and with commutative multiplication are the integers. Notably, no inverses exist except for the identity itself and -1. E.g. the inverse of 2 would be 1/2 which is not in the set.

A polynomial ring is another example with the same properties as the integers.

The simplest non-commutative ring that is not a field is the set of all 2x2 matrices of real numbers:Note that $GL(n)$ is not a ring because you can by addition reach the zero matrix.

- we know that 2x2 matrix multiplication is non-commutative in general
- some 2x2 matrices have a multiplicative inverse, but others don't

Two ways to see it:

- a ring where inverses exist
- a field where multiplication is not necessarily commutative

A ring where multiplication is commutative and there is always an inverse.

A field can be seen as an Abelian group that has two group operations: addition and multiplication, and they are compatible (distributive property).

Basically the nicest, least restrictive, 2-operation type of algebra.

Examples:

One of the defining properties of algebraic structure with two operations such as ring and field:
This property shows how the two operations interact.

$a(b+c)=ab+ac$

A convenient notation for the elements of $GF(n)$ of prime order is to use integers, e.g. for $GF(7)$ we could write:
which makes it clear what is the additive inverse of each element, although sometimes a notation starting from 0 is also used:

$GR(7)={−3,−2,−1,0,1,2,3}$

$GR(7)={0,1,2,3,4,5,6}$

For fields of prime order, regular modular arithmetic works as the field operation.

For non-prime order, we see that modular arithmetic does not work because the divisors have no inverse. E.g. at order 6, 2 and 3 have no inverse, e.g. for 2:
we see that things wrap around perfecly, and 1 is never reached.

$0×2=01×2=22×2=43×2=04×2=25×2=4$

For non-prime prime power orders however, we can find a way, see finite field of non-prime order.

There's exactly one field per prime power, So all we need to specify a field is give its order, notated e.g. as $GF(n)$.

Every element of a finite field satisfies $x_{order}=x$.

It is interesting to compare this result philosophically with the classification of finite groups: fields are more constrained as they have to have two operations, and this leads to a much simpler classification!

As per classification of finite fields those must be of prime power order.

Video 3. "Finite fields made easy by Randell Heyman (2015)" at youtu.be/z9bTzjy4SCg?t=159 shows how for order $9=3×3$. Basically, for order $p_{n}$, we take:For a worked out example, see: GF(4).

- each element is a polynomial in $GF(p)[x]$, $GF(p)[x]$, the polynomial ring over the finite field $GF(p)$ with degree smaller than $n$. We've just seen how to construct $GF(p)$ for prime $p$ above, so we're good there.
- addition works element-wise modulo on $GF(p)$
- multiplication is done modulo an irreducible polynomial of order $n$

Ciro Santilli tried to add this example to Wikipedia, but revert, so here we are.

This is a good first example of a field of a finite field of non-prime order, this one is a prime power order instead.

$4=2_{2}$, so one way to represent the elements of the field will be the to use the 4 polynomials of degree 1 over GF(2):

- 0X + 0
- 0X + 1
- 1X + 0
- 1X + 1

Note that we refer in this definition to anther field, but that is fine, because we only refer to fields of prime order such as GF(2), because we are dealing with prime powers only. And we have already defined fields of prime order easily previously with modular arithmetic.

Over GF(2), there is only one irreducible polynomial of degree 2:

$X_{2}+X+1$

Addition is defined element-wise with modular arithmetic modulo 2 as defined over GF(2), e.g.:

$(1X+0)+(1X+1)=(1+1)X+(0+1)=0X+1$

Multiplication is done modulo $X_{2}+X+1$, which ensures that the result is also of degree 1.

For example first we do a regular multiplication:

$(1X+0)×(1X+1)=(1×1)X_{2}+(1×1)X+(0×1)X+(0×1)=1X_{2}+1X+0$

Without modulo, that would not be one of the elements of the field anymore due to the $1X_{2}$!

TODO show how taking a reducible polynomial for modulo fails. Presumably it is for a similar reason to why things fail for the prime case.

A vector field with a bilinear map into itself, which we can also call a "vector product".

Note that the vector product does not have to be neither associative nor commutative.

Examples: en.wikipedia.org/w/index.php?title=Algebra_over_a_field&oldid=1035146107#Motivating_examples

- complex numbers, i.e. $R_{2}$ with complex number multiplication
- $R_{3}$ with the cross product
- quaternions, i.e. $R_{4}$ with the quaternion multiplication

An algebra over a field where division exists.

Notably, the octonions are not associative.