[proofplan] We prove both directions separately. For $n \le 4$, we exhibit explicit subnormal series with abelian quotients: $S_1$ and $S_2$ are abelian, $S_3$ has the series $\{e\} \trianglelefteq A_3 \trianglelefteq S_3$, and $S_4$ has the series $\{e\} \trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4$ where $V_4$ is the Klein four-group. For $n \ge 5$, we use the fact that $A_n$ is simple and non-abelian, which forces any subnormal series through $A_n$ to have a non-abelian quotient, obstructing solvability. [/proofplan] [step:Verify solvability of $S_1$ and $S_2$] The group $S_1 = \{e\}$ is trivial, hence abelian, hence solvable. The group $S_2 = \{e, (1\;2)\}$ has order $2$, hence is abelian, hence solvable. [/step] [step:Verify solvability of $S_3$ via the subnormal series $\{e\} \trianglelefteq A_3 \trianglelefteq S_3$] The alternating group $A_3$ consists of the even permutations in $S_3$. Since $A_3$ has index $[S_3 : A_3] = 2$, it is normal in $S_3$. The group $A_3$ has order $3!/2 = 3$, so $A_3 \cong \mathbb{Z}/3\mathbb{Z}$ is cyclic, hence abelian. The quotient $S_3 / A_3$ has order $2$, hence is abelian. The subnormal series $\{e\} \trianglelefteq A_3 \trianglelefteq S_3$ has abelian factors $A_3 / \{e\} \cong \mathbb{Z}/3\mathbb{Z}$ and $S_3 / A_3 \cong \mathbb{Z}/2\mathbb{Z}$. Therefore $S_3$ is solvable. [guided] Recall that a group is solvable if it admits a subnormal series whose successive quotients are all abelian. For $S_3$, the natural candidate is the alternating group $A_3$, which consists of the three even permutations $\{e, (1\;2\;3), (1\;3\;2)\}$. Why is $A_3 \trianglelefteq S_3$? Any subgroup of index $2$ in a group is automatically normal: for any $g \in G$, the left coset $gH$ is either $H$ (if $g \in H$) or the unique other coset $G \setminus H$ (if $g \notin H$), and likewise for right cosets, so $gH = Hg$. Since $[S_3 : A_3] = |S_3|/|A_3| = 6/3 = 2$, $A_3$ is normal. Now $A_3$ has order $3$, which is prime, so $A_3 \cong \mathbb{Z}/3\mathbb{Z}$ is cyclic and in particular abelian. The quotient $S_3/A_3$ has order $|S_3|/|A_3| = 2$, so $S_3/A_3 \cong \mathbb{Z}/2\mathbb{Z}$, also abelian. The series $\{e\} \trianglelefteq A_3 \trianglelefteq S_3$ has abelian successive quotients $A_3/\{e\} \cong \mathbb{Z}/3\mathbb{Z}$ and $S_3/A_3 \cong \mathbb{Z}/2\mathbb{Z}$, so $S_3$ is solvable. [/guided] [/step] [step:Verify solvability of $S_4$ via the subnormal series $\{e\} \trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4$] Define the Klein four-group $V_4 = \{e, (1\;2)(3\;4), (1\;3)(2\;4), (1\;4)(2\;3)\} \le A_4$. We verify that $V_4 \trianglelefteq A_4$: $V_4$ consists of the identity and all products of two disjoint transpositions in $S_4$. Since conjugation in $S_4$ preserves the cycle type of a permutation, and the three non-identity elements of $V_4$ are the only elements of $S_4$ with cycle type $(2,2)$, $V_4$ is closed under conjugation by any element of $S_4$, and in particular by elements of $A_4$. So $V_4 \trianglelefteq A_4$. The quotients are: - $V_4 / \{e\} \cong V_4 \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, which is abelian. - $A_4 / V_4$ has order $|A_4|/|V_4| = 12/4 = 3$, so $A_4 / V_4 \cong \mathbb{Z}/3\mathbb{Z}$, which is abelian. - $S_4 / A_4$ has order $2$, so $S_4 / A_4 \cong \mathbb{Z}/2\mathbb{Z}$, which is abelian. All successive quotients are abelian, so $S_4$ is solvable. [guided] For $S_4$, we need to refine the series $\{e\} \trianglelefteq A_4 \trianglelefteq S_4$ further, because $A_4$ is not abelian (it has order $12$). The key is to find a normal subgroup of $A_4$. Define $V_4 = \{e, (1\;2)(3\;4), (1\;3)(2\;4), (1\;4)(2\;3)\}$. This is the set of all permutations in $S_4$ that are products of two disjoint transpositions, together with the identity. One can check that $V_4$ is a subgroup: for instance, $(1\;2)(3\;4) \circ (1\;3)(2\;4) = (1\;4)(2\;3)$, and each non-identity element is its own inverse. So $V_4$ is closed under composition and inverses. Why is $V_4 \trianglelefteq A_4$? Conjugation in $S_n$ preserves cycle type: if $\sigma$ has cycle type $\lambda$, then $\tau \sigma \tau^{-1}$ has cycle type $\lambda$ for any $\tau \in S_n$. The three non-identity elements of $V_4$ are precisely the elements of $S_4$ with cycle type $(2,2)$. Since conjugation preserves this cycle type, $\tau V_4 \tau^{-1} = V_4$ for every $\tau \in S_4$, and in particular for every $\tau \in A_4$. Therefore $V_4 \trianglelefteq A_4$ (in fact $V_4 \trianglelefteq S_4$). Now we check the successive quotients: - $V_4 / \{e\} \cong V_4$. Since $|V_4| = 4$ and every non-identity element has order $2$, $V_4 \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, which is abelian. - $A_4 / V_4$ has order $|A_4|/|V_4| = 12/4 = 3$. A group of prime order is cyclic, so $A_4/V_4 \cong \mathbb{Z}/3\mathbb{Z}$, which is abelian. - $S_4 / A_4$ has order $24/12 = 2$, so $S_4/A_4 \cong \mathbb{Z}/2\mathbb{Z}$, which is abelian. The subnormal series $\{e\} \trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4$ has all abelian quotients, confirming $S_4$ is solvable. [/guided] [/step] [step:Show $S_n$ is not solvable for $n \ge 5$ using simplicity of $A_n$] Suppose for contradiction that $S_n$ is solvable for some $n \ge 5$. Since subgroups of solvable groups are solvable, $A_n$ would also be solvable. By the [Alternating Groups Are Simple](/theorems/849) theorem, $A_n$ is simple for $n \ge 5$. A simple group is solvable if and only if it is cyclic of prime order (since the only subnormal series of a simple group $G$ are $\{e\} \trianglelefteq G$ and $\{G\}$, and solvability requires $G/\{e\} = G$ to be abelian, which for a simple group means $G$ is abelian and simple, hence cyclic of prime order). Since $|A_n| = n!/2$ and $n \ge 5$, we have $|A_n| = n!/2 \ge 60$. In particular $|A_n|$ is not prime (for instance, $60 = 2^2 \cdot 3 \cdot 5$). So $A_n$ is simple but not cyclic of prime order, hence $A_n$ is not solvable. This contradicts the assumption that $S_n$ is solvable. Therefore $S_n$ is not solvable for $n \ge 5$. [guided] We argue by contradiction. Suppose $S_n$ is solvable for some $n \ge 5$. A fundamental property of solvable groups is that every subgroup of a solvable group is solvable. (Proof sketch: if $\{e\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_k = G$ is a subnormal series with abelian quotients, then $\{e\} = H \cap G_0 \trianglelefteq H \cap G_1 \trianglelefteq \cdots \trianglelefteq H \cap G_k = H$ is a subnormal series for $H$, and each quotient $(H \cap G_{i+1})/(H \cap G_i)$ injects into the abelian group $G_{i+1}/G_i$ by the [Second Isomorphism Theorem for Groups](/theorems/843), hence is abelian.) Since $A_n \le S_n$, this would make $A_n$ solvable. But can $A_n$ be solvable when $n \ge 5$? By the [Alternating Groups Are Simple](/theorems/849) theorem, $A_n$ is a simple group for $n \ge 5$. What does simplicity imply about solvability? A solvable group $G \neq \{e\}$ must have a non-trivial abelian quotient in its subnormal series, which means $G$ has a proper normal subgroup (unless $G$ itself is abelian). A simple group has no proper non-trivial normal subgroups, so a simple solvable group must be abelian. An abelian simple group has no proper non-trivial subgroups, hence is cyclic of prime order. Is $A_n$ cyclic of prime order? We have $|A_n| = n!/2$. For $n = 5$, $|A_5| = 60 = 2^2 \cdot 3 \cdot 5$, which is not prime. For $n > 5$, $|A_n| > 60$, and $n!/2$ is divisible by multiple primes. So $A_n$ is simple and non-abelian for $n \ge 5$, hence not solvable. This contradicts our assumption that $S_n$ is solvable, completing the proof that $S_n$ is not solvable for $n \ge 5$. [/guided] [/step]