Back to the on-screen lesson ·
Groups with no normal subgroup to quotient by: the abelian ones settled in five lines, the class-size argument that proves the alternating group on five letters simple, and what the classification lists.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state what makes a group simple, prove that an abelian simple group has prime order, use conjugacy class sizes to list and rule out candidate normal subgroups, show that the alternating group on five letters is simple, say why a non-abelian simple group is never solvable, and describe what the classification of the finite simple groups provides and what it leaves open.
Jordan-Hölder said every finite group breaks into simple factors, uniquely. That makes the question which groups are simple the question of what the pieces can be — and the last lesson showed that a non-abelian simple group is perfect, so it stops a derived series dead. This lesson looks at the pieces themselves.
$G$ is simple when $G \ne 1$ and its only normal subgroups are $1$ and $G$. A group is perfect when it equals its own derived subgroup; every non-abelian simple group is. The classification of finite simple groups lists them all: the cyclic groups of prime order, the alternating groups $A_n$ for $n \ge 5$, sixteen families of groups of Lie type, and twenty-six sporadic groups belonging to no family.
The definition. $G$ is simple when it is non-trivial and has no normal subgroup other than $1$ and $G$.
The point is not that such a group is small or uncomplicated — it is that no quotient is available, so the standard method of studying a group by passing to a smaller one is unavailable from the start.
The abelian case, settled completely. If $G$ is abelian and simple then every subgroup is normal, so $G$ has no proper non-trivial subgroup at all; any non-identity element generates it, so it is cyclic; and a cyclic group of composite order $n$ has a subgroup for each divisor. So $G$ is cyclic of prime order, and conversely every such group is simple. The abelian simple groups are exactly $\mathbb{Z}_p$ for $p$ prime.
The smallest non-abelian one. $A_5$, of order $60$. The quickest proof uses the class sizes $1, 15, 20, 12, 12$: a normal subgroup is a union of classes containing the identity, so its order is $1$ plus a selection of the others, and none of the possible sums divides $60$ except $1$ and $60$.
Why the five-cycles split into two classes of twelve is worth understanding: in $S_5$ they form one class of twenty-four, but conjugating a five-cycle to another may require an odd permutation, which $A_5$ has not got. The centraliser of a five-cycle in $A_5$ has order $5$, so the class has $60/5 = 12$ elements.
And $A_n$ for all $n \ge 5$. The standard argument shows any non-trivial normal subgroup of $A_n$ contains a three-cycle, and that the three-cycles are all conjugate in $A_n$ and generate it. $A_4$ is the last failure: it has $V_4$.
The consequence for solvability. A non-abelian simple group is perfect, so its derived series never descends, so it is not solvable. Since $A_5 \le A_n \le S_n$ for $n \ge 5$, none of those is solvable either — and that is the group-theoretic fact behind the unsolvability of the quintic.
The classification. Announced complete in 2004, the list is: cyclic of prime order; $A_n$ for $n \ge 5$; sixteen infinite families of Lie type; and twenty-six sporadic groups, the largest of which — the Monster — has $808{,}017{,}424{,}794{,}512{,}875{,}886{,}459{,}904{,}961{,}710{,}757{,}005{,}754{,}368{,}000{,}000{,}000$ elements.
Another way: picture
Every other finite group can be taken apart: find a normal subgroup, pass to the quotient, and study two smaller problems. A simple group refuses the first move. There is no smaller group to escape to, so everything that is going to be learned about it has to be learned from the inside — which is why the classification took a century and why the list, once had, is worth so much.
Another way: steps
To decide whether a finite group is simple: 1. If it is abelian, it is simple exactly when its order is prime. 2. If its order is $p^{a}$ with $a > 1$, it is not: the centre is a proper non-trivial normal subgroup. 3. If its order is $p^{a}q^{b}$, it is not: Burnside's theorem makes it solvable. 4. Otherwise try the Sylow counts and element counting, as in the earlier lesson. 5. If nothing rules it out, compute the conjugacy classes and test which unions containing the identity have order dividing $|G|$.
What it gives. Jordan-Hölder says every finite group is built from simple ones. Knowing all the simple groups therefore bounds what any finite group can be made from, and a great many theorems are now proved by reducing to the simple case and consulting the list. The Feit-Thompson theorem — every group of odd order is solvable — is equivalent to saying every non-abelian finite simple group has even order, which the classification confirms.
What it does not give. It does not list the finite groups. Reassembling pieces is the extension problem, and there are already $49{,}487{,}365{,}422$ groups of order $1024$, essentially all of them built from the single simple group $\mathbb{Z}_2$ repeated ten times. Knowing the atoms is not knowing the molecules.
The shape of the list.
| Family | Count | Example |
|---|---|---|
| Cyclic of prime order | one per prime | $\mathbb{Z}_7$ |
| Alternating $A_n$, $n \ge 5$ | one per $n$ | $A_5$, order $60$ |
| Groups of Lie type | sixteen families | $\mathrm{PSL}_2(7)$, order $168$ |
| Sporadic | twenty-six | the Monster |
The first few orders. $60$, $168$, $360$, $504$, $660$ — every one of them with at least three distinct prime factors, as Burnside's theorem requires.
A caution about the word. Simple is a technical term and carries no suggestion of being easy. The Monster group is simple.
*Reading simple as having no subgroups. It means no normal* subgroups other than the two extremes. $A_5$ has subgroups of eight different orders.
Reading simple as small or easy. The Monster is simple and has more elements than there are atoms in the Earth.
Forgetting that $\mathbb{Z}_p$ is simple. The abelian simple groups are easy to overlook precisely because they are so small, and they are exactly the groups of prime order.
Thinking $A_4$ is simple. It is the last alternating group that is not: $V_4$ is normal in it. The theorem starts at $n = 5$.
Expecting simple groups to have even order. They all do, apart from the cyclic ones of odd prime order — but that is the Feit-Thompson theorem, not an observation.
*Concluding simple from no normal Sylow subgroup.* Normal subgroups need not be Sylow subgroups. The class-size test is the one that checks all of them.
Its conjugacy classes have sizes $1, 3, 4, 4$: the identity, the double transpositions, and the three-cycles split into two classes.
The class sizes.
Candidate normal subgroup orders are $1$ plus selections from $3, 4, 4$: that gives $1, 4, 5, 8, 9, 12$.
Form the candidates.
Of those, $1$, $4$ and $12$ divide $12$. And $1 + 3 = 4$ really is a subgroup — $V_4$ — so $A_4$ is not simple. The candidate test found it.
A genuine normal subgroup.
$168 = 2^{3} \cdot 3 \cdot 7$, three distinct primes, so Burnside's theorem does not apply and simplicity is not ruled out.
Nothing obvious forbids it.
$n_7 \mid 24$ and $n_7 \equiv 1 \pmod 7$, so $n_7 \in \{1, 8\}$; $n_7 = 8$ costs $48$ elements of order $7$, which fits inside $168$.
The counting does not close.
And indeed a simple group of order $168$ exists — the second smallest non-abelian one, the symmetries of the Fano plane. The method's failure was not a gap in the argument.
A real group.
The order has only two distinct prime factors.
Count the primes.
Burnside's $p^{a}q^{b}$ theorem says every such group is solvable.
A general theorem applies.
A solvable group that is simple has one composition factor, which is simple and abelian, hence of prime order — so the group would have prime order, and $p^{2}q$ is not prime. So no: never.
$A_5$ has $60$ elements in five conjugacy classes. Fill in the size of each class, and the order a normal subgroup made of the identity together with just that one class would have.
| Elements in the class | Order of the candidate subgroup | |
|---|---|---|
| The identity | ||
| The double transpositions | ||
| The three-cycles | ||
| The first class of five-cycles | ||
| The second class of five-cycles | ||
| All five classes together |
Is this right: every group of even order fails to be simple?
$G$ is simple and has $12$ elements. How many factors does its composition series have?
Answer:
Select every statement that is true of simple groups.
This task has no paper form; do it on a device.
Put in order the steps of proving a group simple from its conjugacy class sizes.
Number the steps in order (write the number in the box):
Build the proof that an abelian simple group is cyclic of prime order.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$G$ is simple and has $8$ elements. How many factors does its composition series have?
Answer:
You can test a group for simplicity from its conjugacy class sizes and say why the abelian simple groups are exactly those of prime order. Say in your own words why simple does not mean having no subgroups. Next: the same structural questions asked of rings, where factorisation replaces normality.
9. Your turn: is a group of order $p^{2}q$ ever simple, for distinct primes $p$ and $q$?, step 3