The key content previously on this page was moved to the following sections:

Minimum number of elements in a generating set of a group.

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

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

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.

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:

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

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.

Ultimate explanation: math.stackexchange.com/questions/776039/intuition-behind-normal-subgroups/3732426#3732426

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

Intuition:

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.

Two ways to see it:

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

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

Ciro Santilli tried to add this example to Wikipedia, but it was reverted, so here we are, see also: Section "Deletionism on Wikipedia".

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):

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:

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

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:

Without modulo, that would not be one of the elements of the field anymore due to the $1X^2$!

So we take the modulo, we note that:and by the definition of modulo:which is the final result of the multiplication.

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.