How are simple groups the building blocks?
Every finite group could be decomposed as a finite number of extensions of simple groups. What is meant by extension could be read here; this is a different form as for example to decompose solely as a direct product (but a direct product could be seen as a special case of a group extension). By repeated extensions by simple groups you get a so called composition series, and these are unique up to a permutation of their composition factors (the quotient groups of successive groups in a series) by Jördan-Hölder. If you are interested there is also a theory for decompositions based on direct products, see Krull-Schmidt theorem.
This question appeared here already in a more or less different flavour, take a look at these posts and their answers:
How is a group made up of simple groups
The "architecture" of a finite group
Let $G=G_0$ be a finite group.
Consider the set of proper nontrivial normal subgroups of $G_0$. If this set is empty, then $G_0$ is simple. Otherwise, the set is ordered by inclusion and we may choose a maximal element $G_1$.
Note that by the correspondence theorem $G_0/G_1$ must be simple and we obtain a short exact sequence $$1\to G_1\to G_0\to G_0/G_1\to 1.$$ This says that $G_0$ is built out of $G_1$ and $G_0/G_1$ (or, $G_0$ is an extension of $G_0/G_1$ by $G_1$).
Now, suppose that $i\geq1$ and we have constructed $$G_{i}< G_{i-1}<\cdots< G_1< G_0$$ with $G_j\lhd G_{j-1}$ and $G_{j-1}/G_j$ simple for $0\leq j\leq i$.
Consider the set of proper nontrivial normal subgroups of $G_{i}$. If this set is empty, then $G_i$ is simple. Otherwise, the set is ordered by inclusion and we may choose a maximal element $G_{i+1}$. Again, by the correspondence theorem we have that $G_i/G_{i+1}$ is simple and we have an exact sequence $$1\to G_{i+1}\to G_i\to G_i/G_{i+1}\to 1.$$ As before, this means the $G_i$ is an extension of $G_i/G_{i+1}$ (a simple group) by $G_{i+1}$.
Now, since $G$ is finite, this process must terminate and we obtain a sequence $$1=G_{n+1}<G_n<\cdots<G_2<G_1<G_0=G$$ with $G_{i+1}\lhd G_i$ and $G_i/G_{i+1}$ simple (in particular, $G_n$ is simple). This is called a composition series for $G$. The subquotients $G_i/G_{i+1}$ are called the composition factors.
Of course, at each step in the construction of the composition series, we chose a maximal normal subgroup. Evidently, this choice need not be unique and it begs the question as to what extent these simple subquotients characterize $G$. This is the content of the Jordan-Holder Theorem.
Theorem (Jordan-Holder): Let $$1=G_{n+1}<G_n<\cdots<G_2<G_1<G_0=G$$ and $$1=H_{m+1}<H_m<\cdots<H_2<H_1<H_0=G$$ be two composition series for $G$. Then, $m=n$ and the set of composition factors for the two series are the same (except possibly for the order in which they appear).
The proof of this fact is not so hard (though it benefits from diagrams which are not so easy to produce on this site). One considers the set of all finite groups for which the above theorem fails. If this set is non-empty, it contains a group $G$ of minimal order.
Now, this $G$ has composition series $$1=G_{n+1}<G_n<\cdots<G_2<G_1<G_0=G$$ and $$1=H_{m+1}<H_m<\cdots<H_2<H_1<H_0=G$$ Note that both $G_1$ and $H_1$ are smaller groups, so they satisfy the Jordan-Holder theorem by assumption. The idea is to find some relationship between their composition series.
Consider a composition series for $G_1\cap H_1$: $$1=K_{r+1}<K_r<\cdots<K_3<G_1\cap H_1$$ (bear with me on the indexing).
Since both $G_1,H_1\lhd G$ are maximal, $G_1H_1=G$. It follows that $H_1/G_1\cap H_1\cong G/G_1$ is simple so $G_1\cap H_1$ is maximal in $H_1$. Therefore, $$1=K_{r+1}<K_r<\cdots<K_3<G_1\cap H_1<H_1$$ is a composition series for $H_1$. In particular, $r=m$. Similarly, $G_1/H_1\cap G_1\cong G/H_1$ is simple, so $$1=K_{r+1}<K_r<\cdots<K_3<G_1\cap H_1<G_1$$ is a composition series for $G_1$ and $r=n$. Hence, the length is unique.
As for the composition factors, we know that the $G/G_1\cong G_1/(G_1\cap H_1)$ and $G/H_1\cong H_1/(G_1\cap H_1)$. Therefore, \begin{align*} \{G/G_1,G_1/G_2,G_2/G_3,\ldots,G_n/G_{n+1}\}&=\{G/G_1,G_1/(G_1\cap H_1),(G_1\cap H_1)/K_3,\ldots, K_n/K_{n+1}\}\\ &=\{G_1/(G_1\cap H_1),G/H_1,(G_1\cap H_1)/K_3,\ldots,K_n/K_{n+1}\}\\ &=\{G/H_1,G_1/(G_1\cap H_1),(G_1\cap H_1)/K_3,\ldots,K_n/K_{n+1}\}\\ &=\{G/H_1,H_1/H_2,H_2/H_3,\ldots,H_n/H_{n+1}\} \end{align*} This completes the proof.
A Final Word: The composition factors themselves are not enough to determine a group. This is easy to see since $\mathbb{Z}/2\times\mathbb{Z}/2$ and $\mathbb{Z}/4$ have the same composition factors (or $\mathbb{Z}_6$ and $S_3$, or $D_8$ and $Q_8$, etc.). To really understand the group, you need to know not only the composition factors, but also the various extensions $$1\to G_{i+1}\to G_i\to G_i/G_{i+1}\to 1.$$
While there is most likely no hope of giving a complete solution to this problem, it has lead to some pretty interesting mathematics.