A5 as the smallest nonabelian simple group

By [L1], $A_5$ is simple. Multiplication by a transposition pairs the even and odd permutations, so $\mathrm{\mid }{A}_5\mathrm{\mid }=5!\mathrm{/}2=60$. The even permutations $\alpha =(012)$ and $\beta =(234)$ do not commute: $(\alpha \beta )(2)=3$ whereas $(\beta \alpha )(2)=0$. Thus $A_5$ is nonabelian.

Suppose $G$ is nonabelian simple and $\mathrm{\mid }G\mathrm{\mid }<60$. Its order cannot be $1$ or a prime power: in the latter case its nontrivial normal center from [L3] would be all of $G$, making $G$ abelian. For every prime divisor $p$ of its order, [L2] therefore gives $n_p>1$.

The non-prime-power integers between $2$ and $59$, apart from $12,24,30,36,48,56$, are $6,10,14,15,18,20,21,22,26,28,33,34,35,38,39,40,42,44,45,46,50,51,52,54,55,57,58.$ For each integer in this list, take its largest prime divisor $p$. Inspection of the divisors of $m=\mathrm{\mid }G\mathrm{\mid }\mathrm{/}{p}^a$ shows that $1$ is the only divisor congruent to $1$ modulo $p$. Thus [L2] gives $n_p=1$, excluding every order in the list.

A nontrivial conjugation action of a simple group on its $n_p>1$ Sylow subgroups is faithful: the kernel is normal and cannot be all of $G$, because the action is transitive. Thus $G$ embeds in $S_{n_p}$. The sign of this action is trivial, since a nontrivial homomorphism from nonabelian simple $G$ to the group of order two would be injective. Consequently $G$ embeds in $A_{n_p}$. For orders $12,24,36$, [L2] gives $n_3=4$. Since $\mathrm{\mid }{A}_4\mathrm{\mid }=12$, [L4] rules out $24,36$; at order $12$ the image would be all of $A_4$, which is not simple: its identity and three double transpositions form a proper nontrivial subgroup preserved by conjugation. For order $48$, [L2] gives $n_2=3$, contradicting an embedding of $G$ into $A_3$, of order $3$.

For order $30$, [L2] and step 1.2 force $n_5=6$ and $n_3=10$. Distinct subgroups of prime order intersect only in the identity by [L4], so these supply $6\cdot 4+10\cdot 2=44$ distinct nonidentity elements, impossible. For order $56$, the same rules give $n_7=8$ and $n_2=7$. The eight subgroups of order $7$ supply $48$ distinct nonidentity elements. Every Sylow $2$-subgroup has order $8$, and its seven nonidentity elements must be the seven remaining nonidentity elements of $G$. All Sylow $2$-subgroups would therefore coincide, contradicting $n_2=7$.

Steps 1.2–2.2 exclude every order below $60$, while step 1.1 supplies a nonabelian simple group of order $60$. Hence $A_5$ is the smallest nonabelian finite simple group. [step 1.1, step 1.2, step 2.1, step 2.2, step 2.3].