The Alternating group $A_5$ on five elements is simple.

Proof

Recall that $A_5$ has order $60$, so Lagrange's theorem states that any subgroup of $A_5$ has order dividing $60$.

Suppose $H$ is a normal subgroup of $A_5$, which is not the trivial subgroup $\{ e \}$. If $H$ has order divisible by $3$, then by Cauchy's theorem there is a $3$-cycle in $H$ (because the $3$-cycles are the only elements with order $3$ in $A_5$). Because $H$ is a union of conjugacy classes, and because the $3$-cycles [4kv form a conjugacy class in $A_n$ for $n > 4$], $H$ would therefore contain every $3$-cycle; but then it would be the entire alternating group.

If instead $H$ has order divisible by $2$, then there is a double transposition such as $(12)(34)$ in $H$, since these are the only elements of order $2$ in $A_5$. But then $H$ contains the entire conjugacy class so it contains every double transposition; in particular, it contains $(12)(34)$ and $(15)(34)$, so it contains $(15)(34)(12)(34) = (125)$. Hence as before $H$ contains every $3$-cycle so is the entire alternating group.

So $H$ must have order exactly $5$, by Lagrange's theorem; so it contains an element of order $5$ since prime order groups are cyclic.

The only such elements of $A_n$ are $5$-cycles; but the conjugacy class of a $5$-cycle is of size $12$, which is too big to fit in $H$ which has size $5$.