Question in Simplicity of $A_5$ using Conjugacy classes argument There is a proof on wikipedia : https://groupprops.subwiki.org/wiki/A5_is_simple
I was searching of proofs that $A_5$ is simple using conjugacy classes argument as I have recently read the chapter on sylow theorems and simple groups which have theory on conjugacy classes.
But I am unable to follow the proof given in the link:

Why the conjugacy classes size is only $1,12, 20, 15$.

I think as $\mbox{cl}(a)$ divides $|G|$, where $\mbox{cl}(a)$ is conjugacy class of $a$ on $G$. It can be $1,2,3,4,5,6,10,12,15,20,30,60$.
Am I wrong ?

"A normal subgroup must contain the conjugacy class of size $1$, and one or more other conjugacy classes"

Is the statement written just above due to following reason?:  If $N$ is a normal subgroup and let  $n\in \mathbb N$ be any of it's elementa then $gng^{-1}$ ( $g \in  G$) lies in $N$ but it is not necessarily $n$ itself (One of the elements it is true when $g = e$(identity)).
"Thus, the order of any normal subgroup must be a sum of some of these numbers, including the $1$. By Lagrange's theorem, the order must also divide the order of the group."
I understand the statement written just above.
"But no such sum among these numbers divides $60$, other than $1$ and $60$ themselves."
I don't understand it this line.

Can you please help with the questions in above proof ?

 A: I attempt an elementary argument.
Before starting notice this. In any group $G$ the number of conjugates of an element $x$ is $|G:C(x)|$.
[Proof: We need to count coincidences among the  $g^{-1}xg$ as $g\in G$. Well $g_1^{-1}xg_1 = g_2^{-1}xg_2$ if and only if $(g_1 g_2^{-1})^{-1}x(g_1 g_2^{-1})=x$; that is, if and only if $(g_1 g_2^{-1})\in C(x)$. Hence all the elements of $C(x)g$ conjugate $x$ to the same element, and that's the only way coincidences happen.]
Any permutation in $S_5$ can be expressed as a product of disjoint cycles, and so has one of the following forms:
$$
(abcde), 
(abcd)(e), 
(abc)(de),
(abc)(d)(e),
(ab)(cd)(e),
(ab)(c)(d)(e)
(a)(b)(c)(d)(e).
$$
Of these the only ones which are even and so lie in $A_5$ are
$$
(abcde), 
(abc)(d)(e),
(ab)(cd)(e),
(a)(b)(c)(d)(e)
$$
whose orders (as listed) are
$$
5,3,2,1.
$$
The number of elements of each type is
$$
\frac{5!}{5}=24,
\frac{5!}{3\cdot 2!}=20,
\frac{5!}{2\cdot 2 \cdot 1 \cdot 2!}=15,
\frac{5!}{5!}=1. 
$$
Now let us split these into conjugacy classes.
(i) As there are no elements of order $10$ or $15$ the order of $C((abcde))$ is (by Lagrange) $5$, and so each of the $24$ $5$-cycles is conjugate to $|A_5:C((abcde))|=12$ others. That is, there are two conjugacy classes of size $12$.
(ii) As there are no elements of order $6$ or $15$ the order of $C((abc)(d)(e))$ is (by Lagrange) $3$, and so each of the $20$ $3$-cycles is conjugate to $|A_5:C((abc)(d)(e))|=20$ others. That is, there is one conjugacy classes of size $20$.
(iii) As there are no elements of order $6$ or $10$ the order of $C((ab)(cd)(e))$ is (by Lagrange) $2$ or $4$, so that each of the $15$ double transpositions $(ab)(cd)(e)$ is conjugate to $|A_5:C((ab)(cd)(e))|$ elements, that is $30$ or $15$ elements. Clearly it is the latter case which holds and there is one conjugacy class of size $15$.
(iv) The identity $(a)(b)(c)(d)(e)$ is in a class of size $1$.
Now let $N$ be a normal subgroup of $A_5$. Since $n\in N$ implies every conjugate $g^{-1}ng\in N$ we have that $N$ is a disjoint union of pieces whose sizes are $1,15,20,12,12$.
Suppose $N\ne A_5$, $N\ne \{(a)(b)(c)(d)(e)\}$. By Lagrange its order divides $60$ and is strictly less than $60$. Also, $N$ must contain the identity and at least one other conjugacy class so its size is at least $13$. The only candidate orders are then $15,20,30$. As $N$ must contain the class of the identity none of these can be built from pieces of the given sizes.
Hence the only normal subgroups of $A_5$ are the identity subgroup and the whole group.
A: You need to know about cycle structures. $A_5$ is the group of all even permutations on $5$ elements. These look like this:

*

*$()$ - the identity

*$(a\ b)(c\ d)$

*$(a\ b\ c)$

*$(a\ b\ c\ d\ e)$
All elements are of the forms above. Moreoever, conjugatation preserves cycle structure. That is, $g$ and $h^{-1}gh$ have the same cycle structure (e.g. if $g$ is a $3$-cycle, then $h^{-1}gh$ is also a $3$-cycle.)
Now, if you accept the statements above, you can accept that the conjugacy classes of $A_5$ are just the sets of different cycle types and you can count them:

*

*cycle type $()$ - just one

*$3$-cycles - $20$ in total

*$(a\ b)(c\ d)$-cycles - $15$ in total

*This case is a bit more complicated.

This is why the proof you use can immediately discount all the other potential sizes for the conjugacy classes and only consider $1,12,15,20$.
The more complicated case is because $5$-cycles split into two conjugacy classes in $A_5$, so rather being size $24$, there are two classes of size $12$. You should find in your textbook or the internet results on cycle structures in $S_n$ and $A_n$, they will give you more details.
This might help you: pp37-39, http://dec41.user.srcf.net/notes/IA_M/groups.pdf
