Having ruled out many small composite orders, we now reach the first genuinely interesting case. The alternating group $A_5$ has order $60$, and it is the smallest nonabelian simple group. In this lesson, the focus keyword is simple groups of order 60 and A5. The main goal is to understand why a simple group of order $60$ must look like $A_5$. This result connects Sylow counting, group actions on cosets, and the alternating group into one important application. :::definition[Alternating Group A5] Let $S_5$ be the symmetric group on five letters. The group $A_5$ is the alternating group of degree five, consisting of all even permutations in $S_5$. ::: Since $|S_5|=120$ and exactly half of the permutations are even, we have $|A_5|=60$. The group $A_5$ contains identity, $3$-cycles, double transpositions, and $5$-cycles. Its conjugacy class structure is the key reason it is simple. :::theorem The group $A_5$ is simple. ::: :::proof Given that $A_5$ is the alternating group of degree five. To prove that $A_5$ is simple. Let $N$ be a normal subgroup of $A_5$. A normal subgroup is a union of conjugacy classes. The conjugacy classes in $A_5$ have sizes $$\begin{equation*} 1,\quad 15,\quad 20,\quad 12,\quad 12. \end{equation*}$$ These correspond respectively to the identity, the double transpositions, the $3$-cycles, and the two classes of $5$-cycles. Therefore $|N|$ must be a sum of $1$ and some of the numbers $15,20,12,12$. Also, $|N|$ must divide $|A_5|=60$. The possible sums containing $1$ and smaller than $60$ are $$\begin{equation*} 13,\ 16,\ 21,\ 25,\ 28,\ 33,\ 36,\ 45,\ 48. \end{equation*}$$ None of these numbers divides $60$. Hence no proper nontrivial normal subgroup of $A_5$ exists. Therefore the only normal subgroups of $A_5$ are $\{e\}$ and $A_5$. Hence $A_5$ is simple. $\square$ ::: This preview shows the class-union argument in the proof that $A_5$ is simple. A normal subgroup must be a union of conjugacy classes and must contain the identity. Select conjugacy classes and compare the resulting size with the divisors of $60$. The key observation is that no selected proper nontrivial class union has an order that divides $60$. :::scientific-preview[A5 Class Sum Explorer] ::: The normalizer calculation comes from counting conjugates of a Sylow subgroup. The coset action turns a subgroup of index $5$ into a permutation representation, which is the bridge from an abstract simple group of order $60$ to $A_5$. :::theorem Any simple group of order $60$ is isomorphic to $A_5$. ::: :::proof Given that $G$ is a simple group of order $60$. To prove that $G\cong A_5$. By the preceding lemma, $G$ contains a subgroup $H$ of order $12$. Hence $$\begin{equation*} [G:H]=\frac{60}{12}=5. \end{equation*}$$ Let $G$ act on the set of left cosets of $H$ by left multiplication. This action gives a homomorphism $$\begin{equation*} \varphi:G\to S_5. \end{equation*}$$ The action is nontrivial, so $\ker\varphi\neq G$. Since $G$ is simple, we must have $$\begin{equation*} \ker\varphi=\{e\}. \end{equation*}$$ Therefore $\varphi$ is injective, and $G$ is isomorphic to a subgroup $T$ of $S_5$ with $$\begin{equation*} |T|=60. \end{equation*}$$ Consider the sign homomorphism restricted to $T$. If $T$ were not contained in $A_5$, then $T\cap A_5$ would be a subgroup of $T$ of index $2$. A subgroup of index $2$ is normal, so this would give a nontrivial proper normal subgroup of $T$, contradicting the simplicity of $T$. Thus $T\subseteq A_5$. Since $$\begin{equation*} |T|=60=|A_5|, \end{equation*}$$ we get $T=A_5$. Hence $G\cong A_5$. $\square$ ::: :::corollary The group $A_5$ is the smallest nonabelian simple group. ::: :::proof Given that every nonprime composite order less than $60$ does not yield a simple group and that $A_5$ is simple of order $60$. To prove that $A_5$ is the smallest nonabelian simple group. Groups of prime order are cyclic and abelian. The previous non-simplicity tests rule out all nonprime orders below $60$ for nonabelian simple groups. Since $A_5$ is nonabelian, simple, and has order $60$, no smaller nonabelian simple group exists. Hence $A_5$ is the smallest nonabelian simple group. $\square$ ::: :::solved-problem Let $G$ be a simple group of order $168$. Show that $G$ has eight Sylow $7$-subgroups and that the normalizer of a Sylow $7$-subgroup has order $21$. ::: :::solution Let $|G|=168=2^3\cdot 3\cdot 7$. Let $n_7$ be the number of Sylow $7$-subgroups. By Sylow's third theorem, $$\begin{align*} n_7 &\equiv 1 \pmod 7, \\ n_7 &\mid 24. \end{align*}$$ Thus $n_7=1$ or $n_7=8$. Since $G$ is simple, $n_7\neq 1$. Hence $$\begin{equation*} n_7=8. \end{equation*}$$ Let $P$ be a Sylow $7$-subgroup of $G$. The Sylow $7$-subgroups are precisely the conjugates of $P$, so their number is $$\begin{equation*} [G:N_G(P)]. \end{equation*}$$ Therefore $$\begin{equation*} [G:N_G(P)]=8. \end{equation*}$$ It follows that $$\begin{equation*} |N_G(P)|=\frac{168}{8}=21. \end{equation*}$$ Hence $G$ has eight Sylow $7$-subgroups and $|N_G(P)|=21$. $\square$ ::: :::exercise 1. Compute $|A_5|$ from $|S_5|$. 2. Explain why a subgroup of index $2$ is normal. 3. If a simple group $G$ of order $60$ has a subgroup of order $12$, explain why it gives an embedding $G\to S_5$. ::: :::answer 1. Since $|S_5|=5!=120$ and the even permutations form a subgroup of index $2$, $|A_5|=60$. 2. A subgroup of index $2$ has exactly two left cosets and two right cosets, so the left and right cosets agree. 3. The action on the five left cosets gives a homomorphism $G\to S_5$. Since $G$ is simple and the action is nontrivial, the kernel is trivial. ::: :::faq[Frequently Asked Questions] Q: Why does $A_5$ matter in Sylow applications? A: It is the first nonabelian simple group and shows that Sylow methods do not always force non-simplicity. Q: Why does the proof use an action on cosets? A: A subgroup of index $5$ creates a natural permutation representation into $S_5$. Q: Why must the image lie in $A_5$? A: If the image had odd permutations, its intersection with $A_5$ would be a normal subgroup of index $2$, contradicting simplicity. :::