What different ways are there to prove that the group $A_5$ is simple?

I've collected these so far:

draks ...
  • 17,825
  • 7
  • 59
  • 171
  • 3
    You forgot the following proof: If $S\subset A_5$ then look at $\mathcal{S}=\langle S\rangle$. Verify that $g^{-1}\mathcal{S}g\neq \mathcal{S}$ for all $g\not\in \mathcal{S}$. Do this for all subsets $S$ of $A_5$. Done. – user1729 Mar 13 '13 at 10:40
  • 4
    @user1729, all 1152921504606846976 subsets? –  Mar 13 '13 at 14:12
  • 6
    Yup. All of 'em! – user1729 Mar 13 '13 at 14:26
  • @user1729, this would make a good answer if you could show the working. obviously if you don't have time no problem. –  Mar 13 '13 at 14:29
  • 3
    It wouldn't really make a good answer - it is just an application of the definition. It is the most inefficient proof possible! – user1729 Mar 13 '13 at 14:36
  • ahh! I didn't understand until now, haha. –  Mar 13 '13 at 14:37
  • 4
    However, since $60 = 2^2 \times 3 \times 5$, any subgroup of $A_5$ is generated by at most 4 elements (the number of prime divisors, with multiplicity). Checking for (give-or-take) ${60 \choose 4} = 487635$ subsets that the subgroup they generate is either trivial, the whole of $A_5$ or non-normal is definitely doable by computer. I think you can even go down to ${60 \choose 3} = 34220$, if you can show that no subgroup of $A_5$ needs 4 generators, but I haven't proved that yet. – yatima2975 Mar 13 '13 at 15:18
  • @yatima2975: Sure, you can optimize it. It is still the most inefficient proof though! – user1729 Mar 13 '13 at 15:28
  • Another proof is discussed [here](http://math.stackexchange.com/q/473673/28900). – Cameron Buie Aug 22 '13 at 16:55
  • Related: http://math.stackexchange.com/questions/15773 – Watson Aug 19 '16 at 21:29

5 Answers5


I'm very happy to show this proof, which makes use of a technique frequently employed in a recent paper of mine. It's my favorite proof that $A_5$ is not solvable, which as you pointed out in your last bullet proves that $A_5$ is simple.

Definition. For $n\in \mathbb{N}$, denote by $\pi(n)$ the set of prime divisors of $n$. The prime graph of a finite group $G$, denoted $\Gamma_G$, is the graph with vertex set $\pi(|G|)$ with an edge between primes $p$ and $q$ if and only if there is an element of order $pq$ in $G$.

By Lucido (1999), Prop. 1, the complement of the prime graph of a solvable group $G$ is triangle-free. It is obvious from cycle types that $A_5$ contains no elements of order $6,10,$ or $15$, so $\Gamma_{A_5}$ is the empty graph on three vertices. Therefore, $A_5$ is not solvable.

Alexander Gruber
  • 26,937
  • 30
  • 121
  • 202

Here's an argument rather specific to $A_5$. Suppose you prove beforehand that $A_5$ is generated by its elements of order $p$ for each of $p=2,3,5$. Then a nontrivial proper quotient is also generated by its elements of order $p$ for each of $p=2,3,5$ and hence has order divisible by $2\times 3\times 5=30$. So we only have to discard normal subgroups of order 2, which amounts to the easy verification that there is no central element of order 2.

For the "beforehand" statement, the easiest is to check it's generated by 3-cycles (true in $A_n$ for all $n$), and then it is enough to write, in $A_5$, a single 3-cycle as product of 5-cycles, and as product of double transpositions.

(I didn't use it, but a side observation is that if a finite group $G$ is generated by its elements of order $p$ for every prime divisor of $|G|$, then every simple quotient of $G$ is nonabelian.)

Added (Oct 2019): as I said the argument is "rather specific" but still can be used in some other cases.

For the group of order $168=2^3.3.7$ a similar argument works. We also have to check that it is generated by its elements of order $p$ for each of $p=2,3,7$, which implies that every proper normal subgroup has index divisible by $2\times 3\times 7$, i.e. order dividing $4$. and then we have to discard a normal subgroup of order $2$ or $4$. Checking that the center is trivial is easy. The other case would be a non-central subgroup of order $4$, but then the centralizer of the latter would then have index $2$, $3$ or $6$, contradiction.

Next, in the case of $A_6$, the generation step is virtually already done in the $A_5$ case, since the primes are the same. Next a normal subgroup has index divisible by $30$, hence order dividing $12$. So, a minimal nontrivial normal subgroup has order $2$, $3$ or $4$ and again: show that the center is trivial, and if not central, the centralizer has too small index.

For $A_7$ one needs to play with elements of order $7$ in the generation step, and then every proper normal subgroup has order dividing $12$, which is ruled out exactly as in $A_6$.

For $A_8$ the argument shows its limits. While the generation step requires no further effort, the main issue will be to rule out specifically the existence of a nontrivial normal subgroup of order dividing $32$. Of course it can be done but the proof becomes quite inefficient.

  • 16,177
  • 26
  • 41

For an easy proof, we can use the following facts.

F1: If $N$ is a normal subgroup of $G$, then $N$ contains every element which has order coprime to $[G:N]$.

F2: $A_5$ contains $15$ elements of order $2$, $20$ elements of order $3$ and $24$ elements of order $5$.

F3: Suppose $N \trianglelefteq G$ and $|N| = 2$. Then $N$ is central in $G$.

F4: $A_5$ has trivial center (more generally, $A_n$ has trivial center when $n \geq 4$).

Let $N$ be a normal subgroup of $A_5$. By Lagrange's theorem, $|N|$ is a divisor of $60$. We can rule out divisors other than $1$, $2$, and $60$ using F1 and F2. For example if $|N| = 15$, then $N$ would contain the $20$ elements of order $3$ and $24$ elements of order $5$, which is absurd. We can rule out $2$ by using F3 and F4. Thus the only possibility is $|N| = 1$ or $|N| = 60$ which proves that $A_5$ is simple.

The idea of this proof is due to Joseph Gallian. He treats the case $|N| = 2$ differently, but the proof is essentially the same as the one here. You can also use the same method to prove that $\operatorname{PSL}(2, 7)$ is simple.

Mikko Korhonen
  • 23,479
  • 3
  • 54
  • 110

Viewing $A_5$ as the rotational symmetries of an icosahedron, one can easily verify that each conjugacy class generates the group. There are really only three cases, as while there are two conjugacy classes of five cycles, they are evidently power equivalent, so we lump them together. It then suffices to see that any vertex can be carried to any other by rotations through vertex axes, any face can be taken to any other by rotations through face axes, and any edge can be taken to any other by rotations through edge axes (as each respective case clearly has the relevant stabilizers, just transitivity of the action establishes that everything is generated.

But each of those cases is very straightforward: once you're familiar with it, there's no need for even a model!

Of course, you may want to prove that this group of rotations is $A_5$, but using this established simplicity and the action on the compound of five cubes (which I think is beautiful enough to warrant thinking about), that can be established without issue.

Beren Gunsolus
  • 2,412
  • 9
  • 21

In a paper by Chapman called "An elementary proof of the simplicity of the Mathieu Groups $M_{11}$ and $M_{23}$" he proves this theorem:

Let $p$ be a prime, $G$ a transitive subgroup of $S_p$, and $|G| = n = pmr$, where $m \equiv 1 \, (mod \, p)$ and $r < p$ is a prime. Then $G$ is simple.

Actually in the paper he proves that $r$ must be the residue of $\frac{n}{p}$ mod p, which makes it easy to calculate. The simplicity of these two groups are then given as corollaries of this theorem. Actually it also applies to $A_5$.

$\frac{n}{p} = \frac{60}{5} = 12 \equiv 2 \, (mod \, 5)$ and $2$ is prime. So $A_5$ is simple.

  • 2,785
  • 8
  • 15