The strategy is to compute the conjugacy class sizes in $A_5$, then show that no proper non-trivial normal subgroup can exist, since its order must both divide $60$ and be a sum of conjugacy class sizes including $1$. **Step 1: Determine the conjugacy classes of $A_5$.** The [group](/page/Group) $A_5$ has order $60$. The even permutations in $S_5$ have cycle types $(1,1,1,1,1)$, $(2,2,1)$, $(3,1,1)$, and $(5)$. Using the [Orbit-Stabiliser Theorem](/theorems/796) and the [Conjugacy and Cycle Type](/theorems/798) theorem for $S_5$, then checking which $S_5$-classes split in $A_5$ (a class splits when no odd permutation centralises a representative): | Cycle type | Example | Size in $S_5$ | Splits? | Size(s) in $A_5$ | |---|---|---|---|---| | $(1,1,1,1,1)$ | $e$ | $1$ | — | $1$ | | $(2,2,1)$ | $(1\;2)(3\;4)$ | $15$ | No | $15$ | | $(3,1,1)$ | $(1\;2\;3)$ | $20$ | No | $20$ | | $(5)$ | $(1\;2\;3\;4\;5)$ | $24$ | Yes | $12, 12$ | Check: $1 + 15 + 20 + 12 + 12 = 60$. The five conjugacy classes have sizes $1, 15, 20, 12, 12$. **Step 2: Show no proper non-trivial normal subgroup exists.** A normal subgroup $N \unlhd A_5$ must be a union of conjugacy classes (since $gNg^{-1} = N$ implies each conjugacy class meeting $N$ is contained in $N$). So $|N|$ has the form: \begin{align*} |N| = 1 + 15\lambda + 20\mu + 12\alpha + 12\beta, \end{align*} where $\lambda, \mu, \alpha, \beta \in \{0, 1\}$. By [Lagrange's Theorem](/theorems/782), $|N|$ divides $|A_5| = 60$. The divisors of $60$ are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$. Checking all $16$ combinations of $(\lambda, \mu, \alpha, \beta) \in \{0,1\}^4$: the only values of $1 + 15\lambda + 20\mu + 12\alpha + 12\beta$ that divide $60$ are $1$ (all zero) and $60$ (all one). Therefore $|N| = 1$ or $|N| = 60$, meaning $N = \{e\}$ or $N = A_5$. So $A_5$ is simple.