MIT 18.701 — Lecture 15

Proof: If $q \not \equiv 1 \pmod{p}$, then the argument for $pq = 15$ can be copy-pasted verbatim.

Otherwise, if $q \equiv 1 \pmod{p}$, then the argument for $pq = 10$ says $G$ must look like:

However, just as in the $pq = 10$ case, we can constrain what $n$ can be, like so:

To better understand the condition $n^p \equiv 1 \pmod{q}$, recall from Problem Set #4 that the multiplicative group $(\mathbb{F}_q^{\times}, \times)$ is cyclic. Thus, $\{n \in \mathbb{F}_q^{\times} \mid n^p = 1\}$ is a cyclic subgroup; say it's generated by some $n_0 \in \mathbb{F}_q^{\times}$. Then:

So $G$ must be $G_{(n_0^k)}$, with $k \in \{0, 1, \dots, p - 1\}$. It turns out that almost all $k$ yield the same $G_{(n_0^k)}$:

Claim. For any $k \in \{1, 2, \dots, p - 1\}$, we have $G_{n_0} \cong G_{(n_0^k)}$.

Proof: The isomorphism is just via a change of generator $y$. Observe the following identity:

$$yx = x^n y ~ \iff ~ y^kx = x^{(n^k)}y^k.$$

Therefore, we may write:

$$\begin{align*}G_{n_0} = \ & \{ \langle x, y \rangle \mid x^q = y^p = \mathrm{id}, \ yx = x^{n_0}y \} \\ = \ & \{ \langle x, y \rangle \mid x^q = y^p = \mathrm{id}, \ y^k x = x^{(n_0^k)}y^k \} \\ = \ & \{ \langle x, y^k \rangle \mid x^q = (y^k)^p = \mathrm{id}, \ y^kx = x^{(n_0^k)}y^k \} \\ = \ & \{ \langle x, y' \rangle \mid x^q = (y')^p = \mathrm{id}, \ y'x = x^{(n_0^k)}y' \} = G_{(n_0^k)}. ~~~ \square\end{align*}$$

So either $n = 1$ or $n \equiv n_0^k \pmod{q}$ for some $k \in \{1, 2, \dots, p - 1\}$. The former yields $G \cong C_p \times C_q$, whereas the latter yields the semi-direct product $C_q \rtimes C_p$. (For now, just trust that $C_q \rtimes C_p$ is correct.)   $\blacksquare$

Proof: For ease of writing, say $\ell = \nu_p(|G|)$ and $|G| = m p^{\ell}$. Consider the following group action:

Group Action. View $G$ as a group action (via left multiplication) on the set $S$ of all subsets $T \subseteq G$ such that $|T| = p^{\ell}$.

We'll use this group action to help us find a subgroup of order $p^{\ell}$.

Claim 1. There is some subset $U \in S$ such that the orbit $GU$ has size not divisible by $p$.

Proof: This is a counting argument. The number of subsets in $S$ is:

$$|S| = \binom{mp^{\ell}}{p^{\ell}} \equiv \binom{m}{1} \equiv m \not \equiv 0 \pmod{p} ~~~~ \text{ by \href{https://en.wikipedia.org/wiki/Lucas\%27s\_theorem}{Lucas' Theorem}}.$$

If the orbits are $GU_i$, then $\sum |GU_i| = |S| \not \equiv 0 \pmod{p}$, so some $U_i$ must satisfy $|GU_i| \not \equiv 0 \pmod{p}$.   $\square$

We'll now use the subset $U$ to construct a subgroup of order $p^{\ell}$.

Claim 2. The stabilizer $H = \Stab_G(U)$ is a subgroup of $G$ with order $p^{\ell}$.

Proof: Stabilizers are always subgroups, so $H$ being a subgroup of $G$ comes for free.

The Orbit-Stabilizer Theorem says that $|G| = |GU| \cdot |H|$. Since $p^{\ell}$ divides $|G|$, yet $p$ does not divide $|GU|$, this forces $p^{\ell}$ to divide $|H|$.

Finally, $HU = U$ implies $U$ is a union of cosets of $H$, and so $|H|$ divides $|U|$. But $U \in S$, so $|U| = p^{\ell}$ must hold. Thus, $|H|$ divides $p^{\ell}$.

Combining “$p^{\ell}$ divides $|H|$” and “$|H|$ divides $p^{\ell}$” yields $|H| = p^{\ell}$, as desired.   $\square$

And so we're done; our desired Sylow $p$-subgroup is $H = \Stab_G(U)$.   $\blacksquare$

Proof: Let $K$ be a $p$-subgroup of $G$. Consider the following group action:

Group Action. View $K$ as a group action on the set of left cosets $S := \{gH \mid g \in G\}$.

It turns out we can say a lot about this group action:

Claim 1. This group action has an orbit of size $1$. In other words, $K(gH) = gH$ for some coset $gH$.

Proof: Say the orbit sizes are $n_1, n_2, \dots, n_t$. Then we have:

$$n_1 + n_2 + \dots + n_t = |S| = \dfrac{|G|}{|H|} \not \equiv 0 \pmod{p}.$$

Thus, there must exist some $i$ for which $n_i \not \equiv 0 \pmod{p}$. But by the Orbit-Stabilizer Theorem, each $n_i$ is a divisor of $|K| = p^k$. Combining $n_i \not \equiv 0 \pmod{p}$ and $n_i \mid p^k$ forces $n_i = 1$ for some $i$.   $\square$

Well, what does $K(gH) = gH$ mean?

Claim 2. If $K(gH) = gH$, then $K \subseteq gHg^{-1}$ (independent of the context of $K$ and $gH$).

Proof: Consider the following (more familiar) group action:

Group Action. View $G$ as a group action on subsets of itself (via left multiplication).

Then we can interpret $K(gH) = gH$ as:

$$K(gH) = gH ~ \iff ~ K \subseteq \Stab_G(gH) = g\Stab_G(H)g^{-1} = gHg^{-1}. ~~~ \square$$

And Claim 2 says exactly what we want.   $\blacksquare$

Proof: Let $X$ denote the set of all Sylow $p$-subgroups.

Claim 1. We must have that $|X|$ divides $|G|$.

Proof: Consider the group $G$ acting on the set $X$ via conjugation. Sylow #2 says this is a transitive group action. Then the Orbit-Stabilizer Theorem says $|G| = |X| \cdot |\Stab_G(H)|$ for any $H \in X$, so $|X|$ divides $|G|$.   $\square$

Showing $|X| \equiv 1 \pmod{p}$ is trickier. This time, pick any $H_0 \in X$, and consider this group action:

Group Action. View $H_0$ as a group action on the set $X$ via conjugation.
(Explicitly, an element $h_0 \in H_0$ acts on a set $H_1 \in X$ by sending it to $h_0H_1h_0^{-1} \in X$.)

Consider the orbit sizes $n_1, n_2, \dots, n_k$ of this group action. Then Orbit-Stabilizer tells us $n_i$ divides $|H_0| = p^{\nu_p(|G|)}$. In other words, for all $i$, either $p \mid n_i$ or $n_i = 1$.

Claim 2. There is exactly one orbit with size $n_i = 1$. (This orbit, of course, is $\{H_0\}$.)

Proof: Suppose some orbit $\{H'\}$ has size $n_i = 1$. Consider the normalizer $N(H')$, defined by:

$$N(H') := \{n \in G \mid nH'n^{-1} = H'\}$$

Then if $\{H'\}$ is an orbit, we must have $H_0 \subseteq N(H')$. Also, $H' \subseteq N(H')$ (obviously).

The key, now, is to think of $H_0$ and $H'$ as Sylow $p$-subgroups of $N(H')$. By Sylow #2, this means $H_0$ is the conjugate of $H'$ by some $n \in N(H')$; that is, $H_0 = nH'n^{-1}$ for some $n \in N(H')$.

But $nH'n^{-1} = H'$ for all $n \in N(H')$ by definition. So $H_0 = H'$, as desired.   $\square$

Thus, we can take (WLOG) $n_1 = 1$ and $p \mid n_i$ for all $i > 1$, which yields: