I have a doubt concerning my part "4-". And I iwould like to have your feedback on it. So you can jump directly there to read it (and read the other part if you feel that some justification are missing).
Question:
Let $G$ be a normal non-trivial sub group of $A_5$. Prove that $G=A_5$.
My answer:
1-
First let remind the definition of a normal sub group. A sub group $H$ of a group $G$ is said to be normal to $G$ if $\forall g \in G \Rightarrow gHg^{-1} \in H $. In more simple words it means that a normal sub group is "stable" under the action of conjugaison by any element of $G$ .
2-
By definition $A_5$ contains all the permutation that can be expressed by an even number of transposition (or equivalently that contains only an even number of transposition).
Thus $A_5$ only contains permutation of this form: 5-cycles $(a;b;c;d;e)$, 3-cycles $(a;b;c)$, product of two dysjoint 2-cyles $(a;b)(c;d)$ or a 0-cycle that means the identity permutation.
More over let remind to that the product of two disjoints 2-cycles can be written as the product of two 3-cycles. Indeed $(a;b)(c;d)=(a;c)(c;b)(c;b)(c;d)=(a;b;c)(b;c;d)$.
And because all 5-cycles can be expressed as composed of 2-cycles ( $(a;b;c;d;e)=((a;b)(b;c)(c;d)(d;e)$) it can be expressed too as composed of 3-cycles.
3-
We have all ready proved here that all the 3-cycles of $A_n$ are pairwise conjuguate in $A_n$ thus from "2-" we can easilly generalize and writte that all the elements of $A_5$ are pairwise conjuguate.
*!!!EDIT this is wrong but not important for the rest of the prove. thk to the comment of Derek Holt!!! More generally in $A_5$ all elements of the same order are conjugate to each other. *
4-
Let's assume that $G$ is a non trivial normal subgroup so it must contains at least one element $g' \in G$ that is not the permutation $Id$.
(i) In this case by the defintion of a normal sub group $\forall \sigma \in A_n \Rightarrow \sigma g' \sigma^{-1} \in G$
(ii) But on the other hand $\forall \sigma , \sigma' \in A_n , \exists \tau \in A_n $ s.t. $ \sigma' = \tau \sigma \tau^{-1}$.
And if we now take $\sigma= g' \Rightarrow $ then by (i) we have that $ \forall \tau \in A_n \Rightarrow \tau g' \tau^{-1} \in G$ and by (ii) the formula $\tau g' \tau^{-1}$ describe $A_n$ entirely whan we make run $\tau $ over all the possible permutation of $A_n$.
Q.E.D.
I have a doubt concerning my part "4-". And I iwould like to have your feed back on it. Indeed i am not sure if i did not "entangle" myself in the reasonning.
EDIT: More over I ve just noted that I ve made a mistake in the part "4-" while does not using the "3-cycles" property that I ve explained before. I am going to try to correct it.
Thank you.
Writing my proof clearly has helped me to see where I was wrong so let's rewrite "4-".
a-
Let's assume that $G$ is a non trivial normal subgroup so it must contains at least one element $g′ ∈ G$ that is not the identity.
If $g'$ is 3-cyles so we are go to step "b-".
If $g'$ is of the form $\sigma = (ab)(cd); a \neq b \neq c \neq d$ then by definition of $G$ being a normal sub group we must have $(bde) \sigma (bde)^{-1} =(bde) \sigma (edb) \in G $ and after that $(bde) \sigma (edb) \sigma^{-1} =(bdeca) \in G$ as the mutliplication of two elements $ (bde) \sigma (edb) $ and $ \sigma^{-1}$ in $G$. In other words $G$ contains too a 5-cycle.
Now following with the same logic we are going to prove that the fact that $G$ contains a 5-cycle (written "$\tau$" here) has for consequency that $G$ must contain too at least one 3-cycle. Formally with $ \tau=(abcde)$ we have that $(cde) \tau (cde)^{-1}=(cde) \tau (edc) \in G \Rightarrow (cde) \tau (edc) \tau^{-1}=(cda) \in G$ too as if $G$ is a subgroup so $\tau \in G \Rightarrow \tau^{-1} \in G$.
We writte this 3-cycle $ \in G$ as $g''$.
b-
On the other hand we proved here that the 3-cycles of $A_5$ are pairwise conjuguates in $A_5$.
That means that in particular for $g'' \in G $ and for any fixed $\sigma' \in A_5 $ that is 3-cycle we can write that $\exists \tau \in A_5, \sigma' = \tau g'' \tau^{-1} $ and thus by definition of $G$ being a normal group $ \Rightarrow \sigma' \in G$ too.
In other words by using the definition of a normal subgroup all the 3-cycles can be obtained via $g''$.
But as we explained in "2-" all the 3-cycles describe entirely $A_5$.
Q.E.D.