Group Theory, Actions & Sylow Structure

State Sylow's Theorems (I, II, and III) for a finite group $G$ where $|G| = p^k m$ and $\gcd(p, m) = 1$.

1. Sylow I (Existence): $G$ contains at least one subgroup of order $p^k$ (a Sylow $p$-subgroup). (Note: More generally, for any $p^i \mid |G|$, $G$ contains a subgroup of order $p^i$).

2. Sylow II (Conjugacy & Embedding): Any two Sylow $p$-subgroups of $G$ are conjugate. Furthermore, every $p$-subgroup of $G$ is contained in some Sylow $p$-subgroup.

3. Sylow III (Counting): The number of Sylow $p$-subgroups $n_p = |\text{Syl}_p(G)|$ satisfies: i) $n_p \equiv 1 \pmod p$

ii) $n_p \mid m$

iii) $n_p = [G : N_G(P)]$ for any $P \in \text{Syl}_p(G)$.

State the Orbit-Stabilizer Theorem for a group $G$ acting on a set $X$, and write out the Class Equation.

1. Orbit-Stabilizer: For any $x \in X$, $|\text{Orbit}(x)| \cdot |\text{Stab}(x)| = |G|$. If $G$ is finite, $|\text{Orbit}(x)\} = [G : G_x]$. 2. Class Equation (Conjugation Action of $G$ on itself): $|G| = |Z(G)| + \sum_{i=1}^{r} [G : C_G(x_i)]$ where $x_i$ are representatives of the non-central conjugacy classes.

[Qual Trigger] Prove that if $|G| = p^k$ ($k \ge 1$), then $Z(G) \neq 1$ (the center is non-trivial).

1. Use the Class Equation: $|G| = |Z(G)\} + \sum [G : C_G(x_i)]$. 2. Each centralizer $C_G(x_i)$ is a proper subgroup containing $Z(G)$, so its index $[G : C_G(x_i)]$ is a positive power of $p$ ($p^{a_i}$ where $a_i > 0$). 3. Thus, $p$ divides every term in the sum except $|Z(G)|$, implying $p \mid |Z(G)|$. 4. Since $1 \in Z(G)$, $|Z(G)| \ge p > 1$, so $Z(G)$ is non-trivial.

[Qual Trigger] What is the standard modular arithmetic trick to prove a Sylow $p$-subgroup is normal, and why does it work?

1. Check $n_p \equiv 1 \pmod p$ and $n_p \mid m$. 2. If the only divisor of $m$ congruent to $1 \bmod p$ is $1$, then $n_p = 1$. 3. If $n_p = 1$, the unique Sylow $p$-subgroup $P$ is invariant under all inner automorphisms (conjugation by any $g \in G$), hence $P \trianglelefteq G$.

[Qual Trigger] Outline the proof that any group $G$ of order $99 = 3^2 \cdot 11$ is abelian.

1. Sylow $11$: $n_{11} \mid 9$ and $n_{11} \equiv 1 \pmod{11} \implies n_{11} = 1$. Let $P_{11}$ be the unique normal Sylow $11$-subgroup. 2. Sylow $3$: $n_3 \mid 11$ and $n_3 \equiv 1 \pmod 3 \implies n_3 \in \{1, 11\}$. But $11 \not\equiv 1 \pmod 3$, so $n_3 = 1$. Let $P_9$ be the unique normal Sylow $3$-subgroup. 3. Direct Product: $P_9 \cap P_{11} = \{e\}$ and both are normal $\implies G \cong P_9 \times P_{11}$. 4. Groups of order $p^2$ are always abelian ($P_9$ is abelian). Thus $G$ is a direct product of abelian groups, hence abelian.

[Qual Trigger] Prove that a group $G$ of order $p^2 q$ (where $p, q$ are distinct primes) cannot be simple.

1. Apply Sylow III: $n_q \mid p^2 \implies n_q \in \{1, p, p^2\}$. 2. If $n_q = 1$, $Q \trianglelefteq G$ (not simple, done). 3. If $n_q > 1$, then $n_q$ must be $p$ or $p^2$. Since $n_q \equiv 1 \pmod q$, analyze cases: if $q > p$, $n_q \equiv 1 \pmod q$ cannot be $p$ or $p^2$ unless $q \mid p-1$ or similar constraints. 4. Alternatively, count elements of order $q$: each non-trivial subgroup of order $q$ has $q-1$ elements of order $q$, and they intersect trivially. If $n_q = p^2$, counting elements forces the Sylow $p$-subgroup to be unique and normal ($n_p = 1$).

Define a transitive action and a doubly transitive (2-transitive) action. What is the relation to stabilizers?

1. Transitive: An action of $G$ on $X$ is transitive if for any $x, y \in X$, there exists $g \in G$ such that $g \cdot x = y$ (i.e., there is only one orbit: $X$). 2. Doubly Transitive (2-transitive): For any two pairs of distinct points $(x_1, x_2)$ and $(y_1, y_2)$ in $X \times X$, there exists $g \in G$ such that $g \cdot x_1 = y_1$ and $g \cdot x_2 = y_2$. 3. Stabilizer Link: An action of $G$ on $X$ is 2-transitive if and only if for any $x \in X$, the induced action of the stabilizer $G_x$ on $X \setminus \{x\}$ is transitive.

[Qual Trigger] Given a short exact sequence $1 \to A \to B \to C \to 1$, what are the standard Q1 proof patterns regarding centers and normal subgroups?

1. Exactness Property: $\operatorname{Im}(f) = \ker(g)$ at every step. In particular, $A$ embeds normally into $B$ ($A \trianglelefteq B$), and $B/A \cong C$. 2. Center/Centralizer Lemma: If $A \subseteq Z(B)$, then $B/A$ acting on elements or centralizer extensions allows lifting central elements from $C$ back to $B$. 3. Splitting: The sequence splits ($B \approx A \rtimes C$) if there exists a homomorphism $s: C \to B$ such that $g \circ s = \operatorname{id}_C$.

[Qual Trigger] Let $H \neq 1$ be a normal subgroup of a finite group $G$. Prove that $H \cap Z(G) \neq 1$.

1. $G$ acts on $H$ by conjugation: $g \cdot h = ghg^{-1} \in H$ (since $H \trianglelefteq G$). 2. Apply the class equation / orbit-size breakdown to this action on $H$. 3. The fixed points of this action are elements $h \in H$ such that $ghg^{-1} = h$ for all $g \in G$, which is precisely $C_G(H) \cap H = Z(G) \cap H$. 4. The size of $H$ satisfies $|H\} = |Z(G) \cap H| + \sum [H : C_H(x_i)]$. Modulo prime power arguments or orbit size divisions, non-trivial orbits must balance, forcing $|Z(G) \cap H| > 1$.

Define the commutator subgroup $G'$ and state the universal property regarding abelian quotients $G/N$.

1. Definition: $G' = \langle [x, y] \mid x, y \in G \rangle$ where $[x, y] = xyx^{-1}y^{-1}$. 2. Quotient Criterion: For any normal subgroup $N \trianglelefteq G$, $G/N$ is abelian if and only if $G' \subseteq N$. 3. Universal Property: $G'$ is the smallest normal subgroup of $G$ such that the quotient is abelian. Any homomorphism from $G$ to an abelian group factors through $G/G'$.

[Qual Trigger] Let $H \le G$ have index $[G : H] = p$, where $p$ is the smallest prime dividing $|G|$. Prove that $H \trianglelefteq G$.

  1. Consider the left multiplication action of $G$ on the set of cosets $G/H$, giving a permutation representation $\rho: G \to S_p$.
  2. The kernel $\ker(\rho)$ is contained in $H$ because if $g \in \ker(\rho)$, then $g(1H) = 1H \implies g \in H$.
  3. By the First Isomorphism Theorem, $G/\ker(\rho)$ embeds into $S_p$, so $|G/\ker(\rho)| \mid p!$.
  4. If $p$ is the *smallest* prime dividing $|G|$, any factor of $|G/H|$ or index analysis forces $\ker(\rho) = H$, making $H$ a normal subgroup.