How is $C_4$ constructed from the simple factor groups of its decomposition series I often hear that a every finite group is build up from simple groups (specifically the factors in its decomposition series, which are unique by Jordan-Hölder). The cyclic group of order $4$ has a decomposition series
$$ 1 \lhd C_2 \lhd C_4 $$
with simple factors {$C_2$, $C_2$}.
Obviously $C_4 \neq C_2 \times C_2$ is true and $C_4 \neq C_2 \rtimes C_2$ holds as well,  since in this case the semidiret product just coincides with the direct product, after all $Aut(C_2) = \{id\}$. In what way then is $C_4$ made up from those factors? I already read into extension theory a bit, but I don't feel it gives a concrete answer even for this simple example. What I mean by concrete is that for example  the group $D_4$ (of order 8) can be written as $D_4 = (C_2 \times C_2) \rtimes C_2$. Here it is very clear to me how the group is made up from it's simple factors. Perhabs the naive question I'm really asking is something like this: Can we not define some general product $\star$, such that $C_2 \star C_2 = C_4$?
 A: Like you said, $C_4$ is a group extension of $C_2$ by itself, this means that we have a short exact sequence $$1\to C_2\to C_4\to C_2\to 1$$
You also noted that $C_4$ is not a semidirect product. This is due to $C_4$ only has one nontrivial proper subgroup. This also manifests into the short exact sequence, in general $1\to H\to G\stackrel{\pi}{\to} K\to 1$ realizes $G$ as the semidirect product if and only if there is a splitting $s:K\to G$ so that $\pi\circ s=\mathrm{id}_K$. In this case, the sequence is called a split exact sequence.
The nice thing about semidirect product is that you can write down the group operation kind of explicitly. The data $G\cong H\rtimes K$ is equivalent to an action $K\to \mathrm{Aut}(H)$, which tells you how to multiply $h\in H$ and $k\in K$ together. The same is not true for group extension. It is much less explicit to write down how the group operations look like.
Edit: As Arturo Magidin pointed out, one can still try to write it down with a bit more information. Here is a little outline: the set up we need is an abelian group $H$ with $K$-action, with an exact sequence $1\to H\to G\stackrel{\pi}{\to} K\to 1$ so that the action on $H$ by conjugations in $G$ is the same as the $K$-action that we are given (note that $ghg^{-1}=g'hg'^{-1}$ if $\pi(g)=\pi(g')$).
To simplify the exposition, let's assume that $K$ acts trivially on $H$ (this is called a central extension). We can try to mimic what we did for semidirect product and find a section $s$ of $\pi:G\to K$ and identity $G=\bigsqcup_{k\in K} H\times s(k)=H\times K$ as sets and ask how do we multiply $(h,k)\cdot (h',k')$. The difference now is that unlike semidirect product, we can only pick a set theoretic section $s:K\to G$, but not a homomorphism. In that case, because $\pi$ is a homomorphism, we must have $s(k)s(k')\in\pi^{-1}(kk')=H\times s(kk')$ and so we have a function $c:K\times K\to H$ so that $s(k)s(k')=c(k,k')s(kk')$. Again we need this function because $s$ is not a homomorphism, in the case of semidirect product this function is trivial. Now the multiplication law on $G=H\times K$ is given by $(h,k)\cdot (h',k')=(h+h'+c(k,k'),kk')$. The associativity and identity axioms of $G$ will impose some conditions on what the function $c$ should satisfy, for example the cocycle condition: $c(k,k'')-c(kk',k'')+c(k,k'k'')-c(k,k')=0$. In particular this is classified by group cohomology $H^2(K,H)$. For more details, you can check out this note by Brian Conrad.
These all are quite abstract, so it is instructive to put your example in question under this lens. $C_4$ is abelian anyway so $C_2$ is contained in the center, so the group action by conjugation is trivial. There are two cosets of $C_2$ in $C_4$, let's call them $C_2, 1\cdot C_4$. So we are picking set theoretic section $s:C_2\to C_4/C_2$ by $s(0)=0$ and $s(1)=1$. Now we need to determine the cocycle function $c:C_2\times C_2\to C_2$. We have:
(1) $s(0)+s(0)=s(0)\Longrightarrow c(0,0)=0$
(2) $s(0)+s(1)=s(1)+s(0)=s(1)\Longrightarrow c(0,1)=c(1,0)=0$
(3) $s(1)+s(1)= 1+s(0)$ so $c(1,1)=1$ (careful that $2\in C_4$ is the image of $1\in C_2\to C_4$.)
So we can write down the product for $C_4$ in terms of $C_4=C_2\times C_2$ as set as follows: $$(a,b)+(c,d)=\begin{cases}(a+c,b+d), \,\text{if}\,\, b\neq 1\, \text{ or}\,\, d\neq 1\\(a+c+1,b+d),\, \text{if}\,\, b=d=1 \end{cases}$$
One can verify that it is indeed the case under the identification $$\begin{align}(0,0)&\leftrightarrow 0\\(0,1)&\leftrightarrow 1\\(1,0)&\leftrightarrow 2\\ (1,1)&\leftrightarrow 3 \end{align}$$
Of course, we ended up with something that is even more complicated than $C_4$, but that's about as general as one can hope to get for a description of group extensions in general.
A: This is a good question.
I think historically, the hope in light of the Jordan-Holder Theorem, was that we would be able to understand/classify finite groups by doing two things:

*

*Determining and understanding all finite simple groups; and

*Figuring out how to understand all extensions of $K$ by $Q$, at least in terms of our understanding of each of $K$ and $Q$.

Now,  it turned  out that part 1 was hard (as witnessed by the Classification of Finite Simple Groups) but doable; but 2 is really hard.
Now, there is a class of cases where one can actually accomplish part 2 above pretty well, and it includes the case in which the normal subgroup is abelian; and even more so when the normal subgroup is central, which is the case here.
The basic idea is that we need an extra piece of information to construct the multiplication table. Just like we need an extra piece of information to construct a semidirect product $N\rtimes Q$ (namely, a morphism $\theta\colon Q\to\mathrm{Aut}(N)$) now we need an additional piece of information because when $1\to N\to G\to Q\to 1$ is our decomposition, there may be no subgroup of $G$ isomorphic to $Q$ to intersects $N$ trivially (as is the case in your situation, where $N=C_2$, $Q=C_2$, and we want $G$ to be $C_4$).
This is accomplished by the use of cocycles and cohomology. You can find an exposition of this in Rotman's Introduction to the Theory of Groups;  in the 4th Edition, it is Chapter 7.
Briefly: assume that you already have a group $G$ with a normal subgroup $N$ and quotient $G/N = Q$. We can define a function $\ell\colon Q\to G$ with the property that $\pi\ell=\mathrm{id}_Q$, where $\pi$ is the canonical map $G\to G/N=Q$. If we can pick this map to be a group homomorphism then we get a semidirect product. But in general, we will have that $\ell(xy)\neq \ell(x)\ell(y)$ for $x,y\in Q$. So we need a "correction factor" that keeps track of this issue.
Explicitly, given an abelian group $N$ written additively, a group $Q$, a morphism $\theta\colon Q\to \mathrm{Aut}(N)$, we say that a function $f\colon Q\times Q\to N$ is a cocycle if and only if:

*

*$f(1,y)=f(x,1)=0$ for all $x,y\in Q$; and

*$\theta(x)(f(x,y)) - f(xy,z) + f(x,yz) - f(x,y) = 0$, the cocycle identity.
The second condition looks complicated, but it essentially comes from the following: we will define a group operation on $N\times Q$, but this operation will not satisfy $(0,x)\cdot (0,y) = (0,xy)$ in general, because it's not going to be a semidirect product. Instead, it will be given by $(0,x)\cdot (0,y) = (f(x,y),xy)$ (that's the "correction factor"). The identity in 2 above is precisely what is required to make sure that the resulting product is associative.
So given $N$, $Q$, $\theta\colon Q\to\mathrm{Aut}(N)$, and a cocycle $f$, we define a group $G$ which is an extension of $N$ by $Q$, as follows:
The underlying set of $G$ is $N\times Q$, ordered pairs $(a,x)$ with $a\in N$ and $x\in Q$.
The operation $\cdot$ on $G$ is the following:
$$(a,x)\cdot(b,y) = (a+\theta(x)(b) + f(x,y), xy).$$
The cocycle identity ensures this is associative. The  identity element is $(0,1)$. The inverse of $(a,x)$ is
$$(-\theta(x^{-1})(a) - \theta(x^{-1})f(x,x^{-1}),x^{-1}).$$
Note that $N$ lies in the center of the resulting group if and only if $\theta$ is the trivial map, in which case the cocycle identity simplifies to
$$f(y,z)-f(xy,z)+f(x,yz)-f(x,y) = 0.$$
Then the operation on the ordered pairs simplifies to
$$(a,x)\cdot(b,y) = (a+b+f(x,y),xy)$$
and the inverse to
$$(a,x)^{-1} = (-a-f(x,x^{-1}),x^{-1}).$$
Now, in our case we have $N=C_2=\{0,x\}$, and $Q=C_2=\{1,y\}$. The action is trivial, because $N$ is central, so we are in the "easy" situation above.
Define $f\colon Q\times Q\to K$ by $f(1,1)=f(1,y)=f(y,1)=0$ and $f(y,y)=x$. This satisfies the cocycle identity
$$f(b,c)-f(ab,c)+f(a,bc)-f(a,b)=0.$$
Then the resulting group $(K\times Q,\cdot)$ is the cyclic group of order four. Indeed, it is generated by $(0,y)$:
$$\begin{align*}
(0,y)^2 = (0,y)\cdot (0,y) &= (0+0+f(y,y),y^2) = (x,1)\\
(0,y)^3 = (x,1)\cdot(0,y) &= (x+0+f(1,y),y) = (x,y)\\
(0,y)^4 = (x,y)\cdot(0,y) &= (x+0+f(y,y),y^2) = (x+x,1) = (0,1).
\end{align*}
$$
and so we have realized $C_4$ as a group on $K\times Q$ with the help of the cocycle $f$.
