**Proof plan.** We show that any non-trivial normal subgroup $H \trianglelefteq A_n$ (for $n \geq 5$) must equal $A_n$. The proof rests on three claims: $A_n$ is generated by 3-cycles; a normal subgroup containing any 3-cycle equals $A_n$; every non-trivial normal subgroup contains a 3-cycle. **Step 1: $A_n$ is generated by 3-cycles.** [claim: 3-Cycle Generation] Every element of $A_n$ is a product of 3-cycles. [/claim] [proof] Every element of $A_n$ is a product of an even number of transpositions. It suffices to express every product of two transpositions as a product of 3-cycles. For distinct $a, b, c, d$: \begin{align*} (a\ b)(a\ b) &= e, \\ (a\ b)(b\ c) &= (a\ b\ c), \\ (a\ b)(c\ d) &= (a\ c\ b)(a\ c\ d). \end{align*} These cover all cases. [/proof] **Step 2: A normal subgroup containing a 3-cycle equals $A_n$.** [claim: 3-Cycle Implies Full] Let $H \trianglelefteq A_n$. If $H$ contains any 3-cycle $(a\ b\ c)$, then $H = A_n$. [/claim] [proof] It suffices to show every 3-cycle lies in $H$. Any two 3-cycles have the same cycle type, so they are conjugate in $S_n$: there exists $\sigma \in S_n$ with $(a\ b\ c) = \sigma(1\ 2\ 3)\sigma^{-1}$. If $\sigma \in A_n$, then $(1\ 2\ 3) \in \sigma^{-1}H\sigma = H$ by normality. If $\sigma$ is odd, replace it by $\bar{\sigma} = \sigma \cdot (4\ 5)$, which is even (here $n \geq 5$ is used). Since $(1\ 2\ 3)$ and $(4\ 5)$ have disjoint support they commute, so $\bar{\sigma}(1\ 2\ 3)\bar{\sigma}^{-1} = \sigma(4\ 5)(1\ 2\ 3)(4\ 5)\sigma^{-1} = \sigma(1\ 2\ 3)\sigma^{-1} = (a\ b\ c)$. Since $\bar{\sigma} \in A_n$ and $H \trianglelefteq A_n$, we again get $(1\ 2\ 3) \in H$. Since every 3-cycle is in $H$ and 3-cycles generate $A_n$, we conclude $H = A_n$. [/proof] **Step 3: Every non-trivial normal subgroup contains a 3-cycle.** [claim: Contains 3-Cycle] Let $H \trianglelefteq A_n$ with $H \neq \{e\}$. Then $H$ contains a 3-cycle. [/claim] [proof] Let $\sigma \in H$ with $\sigma \neq e$. We consider all possible cycle structures of $\sigma$. *Case (i): $\sigma$ contains a cycle of length $r \geq 4$.* Write $\sigma = (1\ 2\ \ldots\ r)\tau$ with $\tau$ disjoint from $\{1,\ldots,r\}$. Let $\delta = (1\ 2\ 3) \in A_n$. Then $\delta^{-1}\sigma\delta \in H$ by normality, so $\sigma^{-1}\delta^{-1}\sigma\delta \in H$. A direct computation gives $\sigma^{-1}\delta^{-1}\sigma\delta = (2\ 3\ r)$, a 3-cycle. *Case (ii): $\sigma$ contains at least two disjoint 3-cycles.* Write $\sigma = (1\ 2\ 3)(4\ 5\ 6)\tau$. Let $\delta = (1\ 2\ 4)$. Then $\sigma^{-1}\delta^{-1}\sigma\delta = (1\ 2\ 4\ 3\ 6)$, a 5-cycle. By Case (i) applied within $H$, this yields a 3-cycle in $H$. *Case (iii): $\sigma = (1\ 2\ 3)\tau$ with $\tau$ a product of 2-cycles.* Then $\sigma^2 = (1\ 3\ 2)\tau^2 = (1\ 3\ 2)$ (since $\tau^2 = e$), which is a 3-cycle in $H$. *Case (iv): $\sigma = (1\ 2)(3\ 4)\tau$ with $\tau$ a product of 2-cycles.* Let $\delta = (1\ 2\ 3)$ and $u = \sigma^{-1}\delta^{-1}\sigma\delta = (1\ 4)(2\ 3) \in H$. Then let $v = (1\ 5\ 2)u(1\ 2\ 5) = (1\ 3)(4\ 5) \in H$ (using $n \geq 5$). Then $uv = (1\ 4)(2\ 3)(1\ 3)(4\ 5) = (1\ 2\ 3\ 4\ 5) \in H$, a 5-cycle, reducing to Case (i). In all cases, $H$ contains a 3-cycle. [/proof] **Step 4: Simplicity.** Let $H \trianglelefteq A_n$ with $H \neq \{e\}$. By Claim 3, $H$ contains a 3-cycle. By Claim 2, $H = A_n$. Hence $A_n$ has no proper non-trivial normal subgroups, so $A_n$ is simple. $\square$