The Sylow Theorems

$|S_4|=24=2^3\cdot3$, so a Sylow $2$-subgroup has order $8$ and a Sylow $3$-subgroup has order $3$. Let $r=(1\,2\,3\,4)$ and $s=(2\,4)$, permutations of the square's $4$ vertices; the subgroup they generate inside $S_4$ is $\{e,\ (1\,2\,3\,4),\ (1\,4\,3\,2),\ (1\,3)(2\,4),\ (2\,4),\ (1\,3),\ (1\,2)(3\,4),\ (1\,4)(2\,3)\}$ — the dihedral group $D_4$ (Topic 10) realized as $8$ of the $24$ permutations of $S_4$, exactly the order Sylow I promises. For the Sylow $3$-subgroup, any $3$-cycle works: $\langle(1\,2\,3)\rangle=\{e,(1\,2\,3),(1\,3\,2)\}$, order $3$.

$|A_4|=12=2^2\cdot3$, so a Sylow $2$-subgroup has order $4$ and a Sylow $3$-subgroup has order $3$. The Klein four-group $V=\{e,(1\,2)(3\,4),(1\,3)(2\,4),(1\,4)(2\,3)\}$ (Topic 7) is a subgroup of $A_4$ of order $4$: a Sylow $2$-subgroup. And $\langle(1\,2\,3)\rangle=\{e,(1\,2\,3),(1\,3\,2)\}$, order $3$, is a Sylow $3$-subgroup — one of several, as Section 2 will count exactly.

$14=2\cdot7$, so $7^1$ is already the full power of $7$ dividing $14$: by Sylow I, a subgroup of order $7$ must exist, for EVERY group of order $14$, with no search required. Contrast this with the converse of Lagrange's theorem, which fails for divisors that are not prime powers: $A_4$ has order $12$ but no subgroup of order $6$ (Topic 5), because $6=2\cdot3$ mixes two different primes — Sylow I makes no promise there at all.

By Sylow III, $n_3 \equiv 1 \pmod 3$ and $n_3 \mid 4$ (since $|A_4|=12=4\cdot3$), so $n_3 \in \{1,4\}$. Count directly: $A_4$ has $8$ elements of order $3$ (every $3$-cycle on $4$ points), and each Sylow $3$-subgroup $\{e,g,g^2\}$ accounts for exactly $2$ of them. So $n_3 = 8/2 = 4$ — the larger candidate, not $1$. Since $n_3 = [A_4:N_{A_4}(P)] = 4$ and $|A_4|=12$, $|N_{A_4}(P)| = 3 = |P|$: this Sylow $3$-subgroup is self-normalizing.

By Sylow III, $n_2 \equiv 1 \pmod 2$ (odd) and $n_2 \mid 3$ (since $|S_4|=24=8\cdot3$), so $n_2 \in \{1,3\}$. The Sylow $2$-subgroup $D_4=\langle(1\,2\,3\,4),(2\,4)\rangle$ found in Section 1 is self-normalizing: its normalizer in $S_4$ has order $8$, equal to $|D_4|$ itself, so $n_2 = [S_4:N_{S_4}(D_4)] = 24/8 = 3$ — the larger candidate again. The three Sylow $2$-subgroups correspond to the three ways to pair up $\{1,2,3,4\}$ into two blocks of $2$.

$36=2^2\cdot3^2$, so a Sylow $3$-subgroup has order $9$ and $m=4$. By Sylow III, $n_3 \equiv 1 \pmod 3$ and $n_3 \mid 4$; the divisors of $4$ are $1,2,4$, and checking each mod $3$ gives $1,2,1$ — so only $1$ and $4$ are $\equiv 1 \pmod 3$. The answer is no: $n_3=2$ is not a divisor of $4$ at all, and even among the divisors of $4$, $2$ fails the congruence too.

$56=2^3\cdot7$. By Sylow III, $n_7 \equiv 1 \pmod 7$ and $n_7 \mid 8$; the divisors of $8$ are $1,2,4,8$, and only $1,8$ are $\equiv 1 \pmod 7$. If $n_7=1$, that Sylow $7$-subgroup is normal and $G$ is not simple. If instead $n_7=8$: the $8$ Sylow $7$-subgroups have prime order $7$, so any two intersect only in $\{e\}$ (a common nonidentity element would generate both, by Lagrange inside a group of prime order), giving $8\cdot(7-1)=48$ distinct elements of order $7$. That leaves only $56-48=8$ elements for everything else, including the identity — exactly the order of a Sylow $2$-subgroup, so that leftover set IS the unique Sylow $2$-subgroup: $n_2=1$, and $G$ is not simple either way.

$12=2^2\cdot3$. By Sylow III, $n_3 \equiv 1 \pmod 3$ and $n_3 \mid 4$, so $n_3 \in \{1,4\}$ (Section 2). If $n_3=1$, that Sylow $3$-subgroup is normal, done. If $n_3=4$: four Sylow $3$-subgroups of prime order $3$ pairwise meet only in $\{e\}$, giving $4\cdot(3-1)=8$ elements of order $3$, leaving exactly $12-8=4$ elements for a Sylow $2$-subgroup (order $4$) — forcing it to be unique, $n_2=1$. Either way $G$ has a normal subgroup. $A_4$ realizes the second case exactly: $n_3=4$ (Section 2) and its unique Sylow $2$-subgroup is the normal $V=\{e,(1\,2)(3\,4),(1\,3)(2\,4),(1\,4)(2\,3)\}$ from Section 1.

$30=2\cdot3\cdot5$. By Sylow III, $n_5\equiv1\pmod5$ and $n_5\mid6$, giving $n_5\in\{1,6\}$; and $n_3\equiv1\pmod3$ and $n_3\mid10$, giving $n_3\in\{1,10\}$. Suppose toward a contradiction that NEITHER Sylow subgroup is normal, i.e. $n_5=6$ and $n_3=10$. Sylow $5$- and $3$-subgroups have prime order, so distinct ones meet only in $\{e\}$: this gives $6\cdot4=24$ elements of order $5$ and $10\cdot2=20$ elements of order $3$, totaling $44$ nonidentity elements — but $G$ has only $30-1=29$ nonidentity elements to spend. Since $44>29$, the assumption is impossible: at least one of $n_5,n_3$ must equal $1$, and $G$ is not simple.

$15=3\cdot5$, $p=3<q=5$. Since $q-1=4$ and $3\nmid4$, there is no nontrivial twist: every group of order $15$ is $\mathbb Z_3\times\mathbb Z_5\cong\mathbb Z_{15}$ (Topic 8, since $\gcd(3,5)=1$) — cyclic, and the only one.

$6=2\cdot3$, $p=2<q=3$. Here $q-1=2$ and $2\mid2$, so a twist DOES exist: besides $\mathbb Z_6$, there is a non-abelian group of order $6$. It is $S_3$ (Topic 1), where the Sylow $3$-subgroup $\langle(1\,2\,3)\rangle$ is normal and a transposition conjugates it by inversion — $(1\,2)(1\,2\,3)(1\,2)=(1\,3\,2)=(1\,2\,3)^{-1}$ — exactly the order-$2$ automorphism of $\mathbb Z_3$ that $2\mid(3-1)$ predicts.

$35=5\cdot7$, $p=5<q=7$. Does a non-abelian group of order $35$ exist? Check $q-1=6$: since $5\nmid6$, the answer is no — every group of order $35$ is $\mathbb Z_5\times\mathbb Z_7\cong\mathbb Z_{35}$, the unique group of that order.

$|A_5|=60$. Sorting its elements by cycle type: $1$ identity; $15$ of type $(2,2,1)$ (double transpositions); $20$ of type $(3,1,1)$ ($3$-cycles); and $24$ of type $(5)$ ($5$-cycles), split into TWO conjugacy classes of $12$ each rather than one class of $24$ (a $5$-cycle and its square are not conjugate inside $A_5$, only inside the bigger $S_5$). So the class equation is

A normal subgroup $N\trianglelefteq A_5$ is a union of whole conjugacy classes including $\{e\}$, with $|N|$ dividing $60$ (Lagrange). Check every way to add some of $15,20,12,12$ to the $1$ from $\{e\}$: of the $2^4=16$ subsets of $\{15,20,12,12\}$, only the empty one ($1+0=1$) and the full one ($1+15+20+12+12=60$) give a sum that divides $60$ — all $14$ other, proper nonempty combinations give a sum between $1$ and $60$ that fails to divide $60$. So the only normal subgroups of $A_5$ are $\{e\}$ and $A_5$ itself: $A_5$ is simple.

For $p=5$: $m=12$, and $n_5\equiv1\pmod5$, $n_5\mid12$ gives $n_5\in\{1,6\}$; counting directly, the $24$ elements of order $5$ split into groups of $4$, giving $n_5=24/4=6$. For $p=3$: $m=20$, $n_3\in\{1,4,10\}$; the $20$ elements of order $3$ split into groups of $2$, giving $n_3=20/2=10$. For $p=2$: $m=15$, $n_2\in\{1,3,5,15\}$ (odd divisors of $15$); direct construction gives $n_2=5$. Every single count is bigger than $1$ — no Sylow subgroup is ever normal in $A_5$ — exactly consistent with simplicity, but, as Section 3 warned, this alone never PROVES it; the class-equation argument above is what actually does.