The strategy is to compute the conjugacy class sizes in $A_5$, then use the [Normal Iff Union of Conjugacy Classes](/theorems/802) theorem to show no proper nontrivial normal subgroup exists. **Step 1: Conjugacy classes of $A_5$.** The even permutations in $S_5$ have cycle types $(1,1,1,1,1)$, $(2,2,1)$, $(3,1,1)$, and $(5)$. The conjugacy class sizes in $S_5$ for these types are $1$, $15$, $20$, and $24$ respectively. To determine which classes split in $A_5$, we check whether each class contains an odd permutation in its centraliser $C_{S_5}(\sigma)$: - Type $(1,1,1,1,1)$: $e$ has centraliser $S_5$, which contains odd permutations. Class does not split. Size in $A_5$: $1$. - Type $(2,2,1)$: e.g. $(1\;2)(3\;4)$ commutes with $(1\;2)$ (odd). Does not split. Size: $15$. - Type $(3,1,1)$: e.g. $(1\;2\;3)$ commutes with $(4\;5)$ (odd). Does not split. Size: $20$. - Type $(5)$: e.g. $(1\;2\;3\;4\;5)$. Its centraliser in $S_5$ is $\langle(1\;2\;3\;4\;5)\rangle$, which consists entirely of even permutations. The class splits into two classes of size $12$ each. So the conjugacy class sizes in $A_5$ are: $1, 15, 20, 12, 12$. **Step 2: Check for normal subgroups.** Suppose $N \unlhd A_5$. By [Normal Iff Union of Conjugacy Classes](/theorems/802), $N$ is a union of conjugacy classes, and must include $\{e\}$ (size $1$). So: \begin{align*} |N| = 1 + 15\lambda_1 + 20\lambda_2 + 12\lambda_3 + 12\lambda_4, \end{align*} where each $\lambda_i \in \{0, 1\}$. By [Lagrange's Theorem](/theorems/782), $|N|$ must divide $|A_5| = 60$. Checking all $16$ combinations of $(\lambda_1, \lambda_2, \lambda_3, \lambda_4) \in \{0,1\}^4$: the only values of $|N|$ that divide $60$ are $|N| = 1$ (all $\lambda_i = 0$) and $|N| = 60$ (all $\lambda_i = 1$). **Step 3: Conclude.** The only normal subgroups of $A_5$ are $\{e\}$ and $A_5$ itself. Therefore $A_5$ is simple.