Solvable and nilpotent groups, normal series and intuition I'm reading Hungerford's algebra and I'm on Nilpotent and solvable groups chapter. Hungerford starts with:

Consider the following conditions on a finite group G:
i)   G is the direct product of its Sylow subgroups
ii)  If m divides |G|, then G has a subgroup of order m
iii) If |G| = mn with (m,n) = 1, then G has a subgroup of order m

This chapter comes after Sylow theorems, where we have seen that if $|G| = p_1^{\alpha_1}p_2^{\alpha_2} \dotsm p_n^{\alpha_n}$ there is a subgroup of order $p_i^k$ for $k \in \{0, 1, ..., \alpha_i\}$, but we can't guarantee more (take for an example $A_4$, it has subgroups of order 3 and 4, but there is no subgroup of order 6).
So the question What kind of groups have nice subgroup structure? isn't that unexpected. 
What was unexpected?

We shall first define nilpotent and solvable groups in terms of certain 
     ''normal series''of subgroups. In case of finite groups, nilpotent groups are   characterized by condition i) and solvable ones by condition iii).

Hungerford first gives definition of nilpotent groups in terms of ascending central series and definition of solvable groups in terms of commutator subgroups and then proves equivalences with i) and iii).
Conceptual jump between i), ii), iii) and definitions involving normal series is a little too big for me. I can't really see how someone could start thinking about i), ii), iii) and come out with normal series.
I don't have a nice way to think about them and I would really be thankful if someone has a nice conceptual view about them, or several eye-opening exercises.
 A: This is a kind of long answer to explain why normal subgroups are so helpful for finding subgroups. Once you learn about composition series and chief series most of this is pretty easy. Derived series and lower central series make for easier homework questions, but if you don't see where they come from, then it might be comforting to know they are just one of many series that work out. I give some pretty casual series below that give equivalent definitions in finite groups.
The dihedral group of order 8 has subgroups of orders 1, 2, and 4. In fact it has normal subgroups of each of these orders, and every subgroup of order 4 has a subgroup of order 1 and 2.
In fact the same is true for any group of order 8. Something similar is true in any group of order $p^n$: it has normal subgroups of order $p^k$ for all $0 \leq k \leq n$ and every subgroup of order $p^k$ contains subgroups of order $p^i$ for all $o \leq i \leq k$.
The normal subgroups feature is particularly helpful since it allows us to replace the group of order $p^n$ by groups of order $p^k$ and $p^{n-k}$. That is, we can replace $G$ by $N$ and $G/N$ where $N$ is a normal subgroup. Subgroups of $N$ are clearly subgroups of $G$ (with small order), and subgroups of $G/N$ are all of the form $H/N$ where the $H$ are subgroups of $G$ (with large order).
So normal subgroups are nice as far as finding more subgroups in smaller groups. A normal series is just a way to break down a group into lots of small pieces. By normal series of length $d$, I'll mean subgroups $N_i$ such that $1=N_d$, $N_0 = G$, each $N_i$ is normal in $G$, and $N_i \leq N_{i-1}$. It turns out that studying how nice the subgroups of $G$ are is very related to how nice the subgroups between $N_i$ and $H=N_{i-1}$ are, when $N_i \leq N_{i-1} \leq G$.
Super
Here is an example: $N \leq H \leq G$ is super-nice if every subgroup between $N$ and $H$ is normal in $G$. Groups with a super-nice normal series are called supersolvable. They have subgroups of every possible order, and in fact every maximal chain of subgroups between any two subgroups has the same length [and these are the only finite groups with that property].
Proving that super-nice groups have all the subgroups of (ii) should be an easy exercise. Proving that the stronger version of (ii) [ that all maximal chains of subgroups have the same length ] is a theorem of Iwasawa. A related theorem of Huppert is that a group in which every maximal subgroup has index a prime is supersolvable (so if it has the “right” large subgroups, then it has all possible subgroups).
Duper
Doing better than super-nice is pretty hard, but it helps to know that being super nice means that $G$ acts like numbers when it acts on $H/N$. $N \leq H$ is super-nice means that $g^{-1}hg = h^k n$ for some number $k$ and some $n \in N$ (and with some mild hypotheses, the $k$ does not depend on $h$). What is the easiest choice of $k$? Well $k=1$ of course.
$N \leq H \leq G$ is super-duper-nice if for every $g\in G$ and $h\in H$, there is some $n \in N$ so that $g^{-1} h g = h n$. Notice $h^k$ is just plain old $h$ here. That's the duper.
A group of order $p^n$ has a super-duper-nice normal series (in fact a super-nice normal series is automatically super-duper-nice because the possible $k$ have to divide $[G:N]$ and be relatively prime to $[H:N]$, but they are both powers of $p$, leaving $k=1$ as the only possibility). A group that is a direct product of (its Sylow) $p$-groups has a super-duper-nice normal series. It turns out that a finite group with a super-duper-nice normal series is also a direct product of its Sylow $p$-subgroups, mostly because direct products tend to have $k=1$. Such groups are called nilpotent.
The exercises here should be the same as in your textbook.
Barely
$N \leq H \leq G$ is barely-nice if every subgroup between $N$ and $H$ is normal in $H$ (rather than $G$). This is still enough to get lots of subgroups but some are “missing:” if $G$ is a product of three primes, like $2 \cdot 2 \cdot 3$ then you can get a chain of subgroups of the right length, like order $1 < 2 < 2\cdot 2 < 2\cdot 2 \cdot 3$, but maybe you can't get other chains of the right length, $1 < 2 < 2\cdot 3 < 2 \cdot 3 \cdot 2$ is impossible in $A_4$. Sylow theorems help out to let you get as many primes as you want all in a row, but once you change primes you can't necessarily change back. A group is called solvable if it has a barely-nice normal series.
I think the exercises here are a bit more difficult. You should be able to prove that you can get a chain of subgroups in the form of (iii) without much trouble [ using one more prime power in each larger subgroup ], but proving you can choose the order of the primes is a little harder (1928 result of Hall). Proving that if you can do this the group is solvable is another result of Hall, but quite a bit harder.
Naughty
A group is said to be insoluble (or naughty) if it doesn't even have a barely-nice normal series. These groups are crazy and missing tons of subgroups. That's good in a way, since too many subgroups means there is too much to look at.
All finite groups have their subgroups of prime order. If $p$ divides $|G|$, then $G$ has a subgroup of order $p$. So small subgroups just tend to exist. It's the big ones that can vanish. If $G$ is even barely-nice and $p$ divides $|G|$, then $G$ has a subgroup of order $|G|/p^k$ for some $k$. For $G$ super-nice, $G$ has a subgroup of order $|G|/p$.
However, the group $S_{23}$ is pretty messed up. For $p=2$ and $p=23$ it has subgroups of order $|G|/p$. However, none of the other primes $p$ that divide its order (to be clear, that is $p=3,5,7,11,13,17,19$, a LOT of primes) even have subgroups of order $|G|/p^k$ for any positive integer $k$. In fact, other than $|G|/2$ and $|G|/23$ the next biggest subgroup has order $|G|/253$ and then $|G|/1771$. Not $|G|/3$, not $|G|/15$, not even a measly $|G|/200$.
In some sense this is no surprise, $S_{23}$ is almost simple, it has only one proper non-identity normal subgroup $A_{23}$. No normal subgroups means no real reason to have other subgroups. Well Sylow subgroups, and their normalizers. A few exceptional subgroups creep by, but the huge variety of subgroups forced into existence in the super-duper nice groups just isn't here.
