I am currently trying to understand the following proof that $A_5$ is a simple group. I only have the proof in German, so the following is only my translation.

I really want to understand this proof. I have seen other proof (e.g. in this question or on subwiki), but I would like to understand the one below. So I wrote

Theorem: $A_5$ is simple

To show: $A_5$ has only $A_5$ and $\{Identity\}$ as normal groups


The number of elements in $A_n$ is $\frac{n!}{2}$, so $|A_5| = 60 = 2^2 \cdot 3 \cdot 5$.

The 3- and 5- Sylowgroups are isomorph to $\mathbb{Z}/3 \mathbb{Z}$ and $\mathbb{Z}/5 \mathbb{Z}$, so they are cyclic. With the Sylow-theorems we know that:

  • $60 = 3^1 \cdot 20$
    • The number of 3-Sylow groups of $A_5$ divides 20
    • The number of 3-Sylow groups of $A_5$ is of the form $1+k \cdot 3$ with $k \in \mathbb{N}_0$
    • $\Rightarrow$ The number of 3-Sylow groups of $A_5$ is in $\{1,4,10\}$
  • $60 = 5^1 \cdot 12$
    • The number of 5-Sylow groups of $A_5$ divides 12
    • The number of 5-Sylow groups of $A_5$ is of the form $1+k \cdot 5$ with $k \in \mathbb{N}_0$
    • $\Rightarrow$ The number of 3-Sylow groups of $A_5$ is in $\{1,6\}$
  • $60 = 2^2 \cdot 15$
    • The number of 2-Sylow groups of $A_5$ divides 15
    • The number of 2-Sylow groups of $A_5$ is of the form $1+k \cdot 2$ with $k \in \mathbb{N}_0$
    • One 2-Sylowgroup of $A_5$ is the Klein four group: $V_4 = \{(1~2)(3~4), (1~3)(2~4), (1~4,2~3), Id\}$
    • $\Rightarrow$ The number of 2-Sylow groups of $A_5$ is in $\{1,3,5,15\}$

Question 1: Somehow it seems to be clear that the number of 3-Sylow groups is 10, the number of 5-Sylowgroups is 6 and the number of 2-Sylowgroups is 5. Why is this clear? Could you explain that to me?

$A_5$ is generated by 3-cycles and thus also by 5-cycles, because: $$(1~2~3~4~5) \circ (1~3~2~5~4) = (1~4~2)$$

Let $N \lhd A_5$ be a normal subgroup with more than one element. If $\#N | p, \;\;\; p \in \{3,5\}$, then $N$ has a $p$-Sylow group of $A_5$.

This means, $N$ has to have all $p$-Sylow groups (thus all $p$ cycles). This means $N=A_5$

Question 2: Why can't $N$ have only one $p$-Sylow group of $A_5$ and still be a normal subgroup?

And the cardinality $\# N$ cannot be a power of two as the center $Z(A_5) = \{Id\}$. This means $|N| > 2$.

Question 3: I don't understand how the center is related to the number of elements of $N$.

$N$ can't be a 2-Sylow group as all 2-Sylow groups are conjugated.

Question 4: I know that all 2-Sylow groups are conjugated because of Sylow theorems. But why does this mean that $N$ can't be a 2-Sylow group?

It follows that $N = A_5 \blacksquare$

Question 5: Why does this follow? Why does $N$ have to be a p-Sylow group?

Notation and definitions

I'm not sure if I used the correct English words. So here are some definitions:

Let $G$ be a group. The center is defined as $\mathrm Z(G):=\{z \in G \mid \forall g \in G : gz=zg\}.$

Let $G$ be a group. The number of elements of $G$ is denoted as $|G|$ or $\#G$.

Let $G$ be a group. $x, y \in G$ are conjugated $:\Leftrightarrow \exists g \in G: x = gxg^{-1}$.

Let $G$ be a group. $G$ is called a $p$-group $:\Leftrightarrow$ the order of each element of $G$ is $p$

  • $\begingroup$ Minor terminology note: it should be "conjugate" rather than "conjuncted". $\endgroup$
    – mdp
    Aug 22, 2013 at 16:20
  • $\begingroup$ @MattPressland: Thank you! I try to improve my English and I've corrected the mistake. $\endgroup$ Aug 22, 2013 at 16:22
  • $\begingroup$ Take a look at en.wikipedia.org/wiki/Sylow_theorems In particular #2. $\endgroup$
    – LASV
    Aug 22, 2013 at 16:24

4 Answers 4


Q1: There are $\frac{5\cdot4\cdot 3}3=20$ three-cycles in $A_5$, and each is in a $3$-Sylow group, whereas each $3$-Sylow group consists of two elements of order three (plus the identity). Hence the number of $3$-Sylow groups is exactly $10$.

Likewise, each $5$-Sylow group contains four elements of order $5$ and there are $\frac{5!}5$ such elements.

Each choice of four numbers in $\{1,2,3,4,5\}$ gives rise to a copy of the Klein 4-group, hence there are $5$ such subgroups.

Q2: As $A_5$ acts transitively on the set of its $p$-Sylow groups by conjugation and a normal subgroup is invariant under conjugation, we have this all-or-nothing situation: If $N$ contains one $p$-Sylow group, it must contain all its conjugates.

Q3: If $\#N=2$ then the nontrivial element must be fix under conjugation, i.e. central. If $\#N=4$ it must be one of those Klein 4-groups and $\#N=8,16,\ldots$ is not possible

Q4: See Q2. Any subgroup is normal iff it is invariant under conjugation.

Q5 We have seen in Q2 that $N=A_5$ if $3|\#N$ or $5|\#N$, hence from $1<\#N|60$ we are left only with $\#N\in\{2,4\}$ to check. This was shown impossible in Q3, Q4.


An answer to Question 1: Count the numbers of cycles of length 2, 3 and 5.

An answer to Question 2: Since all Sylow $p$-subgroups are conjuncted.

An answer to Question 3: If $|N|=4$ then $N$ coincides with a Sylow $2$-group; if $|N|=2$ then the non-identity element of $N$ is central.

An answer to Question 5: "Why does N have to be a p-Sylow group?" -- See Question 2.


Another way to prove this is by looking at the rotational symmetries of an icosahedron (or dodecahedron) - this group is $A_5$. Perhaps there the question of when two elements are conjugate is more easily seen. Have a look at Artin's Algebra for a proof.

  • $\begingroup$ This is the way I like: it's not hard to show any conjugacy class generates the whole group (there's essential 3 cases, as two classes are power equivalent), so you don't even need to do a counting argument! Then use simplicity and the five inscribed cubes (more natural with dodecahedron) to show this rotation group is $A_5$ $\endgroup$ Feb 22, 2021 at 1:43

Others have answered most parts well. I add some details to the answers on Question 2, because those are repeated all over the place.

If $3\mid \#N$, then $N$ has to contain a Sylow $3$-subgroup $P$ that is also a Sylow $3$-subgroup of $A_5$. The group $P$ is conjugate to all the other Sylow $3$-subgroups of $A_5$ in $A_5$. As $N\lhd A_5$, $N$ must contain all the Sylow $3$-subgroups of $A_5$. But then all the 3-cycles are elements of $N$, because they all belong to Sylow $3$-subgroup. The subgroup of $A_5$ generated by all the 3-cycles is... The same argument works, when $5\mid \#N$.


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