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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 6.1 Sylow’s Theorem

We come to a very powerful technique for analyzing finite groups of relatively small order. One aspect of Sylow theory is that it allows us to deduce, in certain special cases, the existence of a unique subgroup of a given order, and thus it allows one to construct a normal subgroup.

Subsection Groups of Prime Order

“The greatest weakness of most humans is their hesitancy to tell others how much they love them while they’re still alive.”
―Optimus Prime

Definition 6.1. \(p\)-group.

For a prime number \(p\text{,}\) a \(p\)-group is a group of order \(p^m\) for some \(m\in \mathbb{Z},m > 0\text{.}\)

Proof.

Let \(g_1,\ldots g_r \in G\) be a list of unique representatives of all of the conjugacy classes of \(G\) of size greater than \(1\) as in Class Equation. Then for each \(i\text{,}\) \(C_G(g_i)\neq G\) so \([G:C_G(g_i)]\neq 1\text{.}\) Since \(1\neq [G:C_G(g_i)] \mid |G|=p^m\text{,}\) it follows that \(p\mid [G:C_G(g_i)]\) for each \(i\text{.}\) From Class Equation we deduce that \(p\mid |Z(G)|\) so, \(|Z(G)|\neq 1\text{.}\)

Qual Watch.

Proving Theorem 6.2 was Part (a) of [cross-reference to target(s) "jan-2017-2" missing or not unique] on the [cross-reference to target(s) "may-2022" missing or not unique] qualifying exam.

Qual Watch.

Proving Part (b) of Theorem 6.3 was [cross-reference to target(s) "june-2020-1" missing or not unique] on the [cross-reference to target(s) "june-2020" missing or not unique] qualifying exam.
Cauchy’s Theorem allows us to conclude the existence of certain elements in a group based solely on its order, which can be a powerful tool in studying the structure of groups.
For example, it can be used to prove the existence of subgroups of a given order in a group, and it is a key ingredient in the proof of Sylow’s Theorems, which provides information about the structure of finite groups.
Cauchy’s Theorem has since been generalized and extended in many ways, and it is now a fundamental result in the theory of finite groups, where it plays a central role in the classification of groups of small order.

Subsection Sylow Subgroups

Definition 6.5. Sylow \(p\)-subgroup.

Let \(G\) be a finite group and \(p\) a prime. Write the order of \(G\) as \(|G| = p^e m\) where \(p \nmid m\text{.}\) A Sylow \(p\)-subgroup of \(G\) is a subgroup \(H \leq G\) such that \(|H| = p^e\text{.}\) That is, a Sylow \(p\)-subgroup of \(G\) is a subgroup whose order is the highest conceivable power of \(p\) according to Lagrange’s Theorem.
We set \(\operatorname{Syl}_p(G)\) to be the collection of all Sylow \(p\)-subgroups of \(G\) and \(n_p=|\operatorname{Syl}_p(G)|\) to be the number of Sylow \(p\)-subgroups.

Remark 6.6.

We allow the case when \(p\nmid |G|\text{,}\) in which case \(e = 0\) and \(G\) has a unique Sylow \(p\)-subgroup, namely \(\{e\}\) which has order \(p^0\text{.}\)

Example 6.7. Sylow Subgroups in \(D_{2n}\).

In \(D_{2p}\) for a prime \(p\text{,}\) \(\langle r \rangle\) is a Sylow \(p\)-subgroup. If \(p > 2\text{,}\) there is only one Sylow \(p\)-subgroup of \(D_{2p}\text{,}\) so \(n_p=1\text{.}\)
In \(D_{2n}\) for \(n\) odd, each of the subgroups \(\langle sr^j \rangle\text{,}\) for \(j = 0, \dots, n-1\) is a Sylow \(2\)-subgroup, so \(n_2=p\text{.}\)

Example 6.8. Sylow Subgroups in \(S_{5}\).

In \(S_5\text{,}\) the Sylow \(5\)-subgroups are the cyclic groups \(\langle \sigma \rangle\) for any five cycle \(\sigma\text{,}\) so \(n_5=6\) because there are \(24\) five cycles, but there are four of these in every Sylow \(5\)-subgroup. The Sylow \(3\)-subgroups are the cyclic groups \(\langle \sigma \rangle\) for any three cycle \(\sigma\text{,}\) so \(n_3=10\) because there are \(20\) three cycles, but there are two of these in every Sylow \(3\)-subgroup.
A Sylow \(2\)-subgroup of \(S_5\) is any subgroup of order \(8\text{.}\) For example \(\langle (1 \, 4)(2 \, 3) , (1 \, 2 \, 3 \, 4) \rangle\) is a Sylow \(2\)-subgroup. There are many others.

Subsection Sylows-Theorem

“What makes me unique is that I’m normal.”
―Scotty McCreery

Remark 6.10.

As of now, for all we know, this might be the action on the empty set, but we’ll see in a bit that the set of Sylow \(p\)-subgroups is in fact not empty.
Sylow Theory is all about understanding this action very well.

Proof.

Since \(P \leq N_G(P)\text{,}\) we have \(Q \cap P \leq Q \cap N_G(P)\text{.}\) For the reverse containment, let \(H = Q \cap N_G(P)\text{.}\) Since \(H \subseteq N_G(P)\text{,}\) \(PH\) is a subgroup of \(G\text{.}\) (Technically, we only proved \(PH\) is a subgroup if \(N_G(P) = G\text{,}\) but the proof applies verbatim provided the weaker condition \(H \subseteq N_G(P)\) holds.) Also
\begin{equation*} |(PH)| = \frac{(|P|) (|H|)}{|(P \cap H)|} \end{equation*}
and since each of \(|P|\text{,}\) \(|H|\text{,}\) and \(|(P \cap H)|\) is a power of \(p\text{,}\) \(PH\) is a \(p\)-subgroup of \(G\text{.}\) But \(P \leq PH\) and \(P\) is a \(p\)-subgroup of largest possible order. So \(P = PH\text{.}\) This proves \(H \leq P\) and thus \(H \leq Q\cap P\text{.}\)

Proof.

  1. Recall that we write \(|G| = p^em\) where \(p \nmid m\text{.}\) We need to prove \(G\) contains a subgroup of order \(p^e\text{,}\) and we proceed by induction on \(|G|\text{.}\)
    If \(|G| = 1\) or, more generally, if \(p \nmid |G|\text{,}\) then \(\{e\}\) is a Sylow \(p\)-subgroup. We may thus assume \(p \mid |G|\text{.}\) We proceed by considering two cases, depending on whether or not \(p\) divides \(|Z(G)|\text{.}\)
    If \(p \mid |Z(G)|\text{,}\) then by Cauchy’s Theorem, there is an element \(z \in Z(G)\) of order \(p\text{.}\) Set \(N = \langle z \rangle\text{.}\) Since \(z \in Z(G)\text{,}\) we have \(N \unlhd G\text{.}\) Since \(|G/N| = p^{e-1}m\text{,}\) by induction \(G/N\) has a subgroup of order \(p^{e-1}\) (i.e. of index \(m\)). By Lattice Isomorphism Theorem, this subgroup corresponds to a subgroup of \(G\) of index \(m\text{,}\) hence of order \(p^e\text{.}\)
    For the second case, assume \(p \nmid |Z(G)|\) and consider the Class Equation for \(G\text{:}\)
    \begin{equation*} |G| = |Z(G)| + \sum_{i=1}^k [G: C_G(g_i)] \end{equation*}
    where \(g_1, \dots, g_k\) are a complete and non-redundant list of non-central conjugacy class representatives. Since \(p \nmid |Z(G)|\) and \(p \mid |G|\text{,}\) we must have \(p \nmid [G:C_G(g_i)]\) for at least one \(i\text{.}\) For this \(i\text{,}\) we have \(|C_G(g_i)|= p^e j\) where \(p \nmid j\text{.}\) Since \(g_i\) is not central, \(|C_G(g_i)| < |G|\) and hence, by induction, \(C_G(g_i)\) contains a subgroup of order \(p^e\text{.}\) But this subgroup is also a subgroup of \(G\text{.}\)
  2. Let \(P\) be a Sylow \(p\)-subgroup and let \(Q\) be any \(p\)-subgroup. Let \(\mathcal S_P\) denote the collection of all conjugates of \(P\text{:}\)
    \begin{equation*} \mathcal S_P = \{ gPg^{-1} \mid g \in G\}. \end{equation*}
    Part (3) tells us the \(\mathcal S_P\) consists of all Sylow \(p\)-subgroups of \(G\text{,}\) but we don’t yet know this. Nonetheless, \(G\) acts (transitively) on \(\mathcal S_P\) by conjugation, and thus \(Q\) also acts on \(\mathcal S_P\) (not necessarily transitively). The key to proving parts (2) and (3) of the Sylow Theorem is to analyse the action of \(Q\) on \(\mathcal S_P\) to establish () and () below.
    Let \(O_1, \dots, O_s\) be the orbits of the action of \(Q\) on \(\mathcal S_P\text{,}\) and for each \(i\) pick a representative \(P_i \in O_i\text{.}\) We have \(\operatorname{stab}_Q(P_i) = \{q \in Q \mid qP_iq^{-1} = P_i\} = Q \cap N_G(P_i) = Q \cap P_i\text{,}\) where the last equation uses the Lemma. By LOIS, we thus have \(|O_i| = [Q: Q \cap P_i]\) and hence
    \begin{gather*} |\mathcal S_P |= \sum_{i=1}^s [Q: Q \cap P_i] \end{gather*}
    This equation holds for any \(p\)-subgroup \(Q\) of \(G\text{.}\) In particular, we can take \(Q = P_1\text{.}\) In this case the first term is \(1\) and, since \(Q \cap P_i = P_1 \cap P_i < P_1 = Q\) for all \(i \ne 1\text{,}\) the remaining terms are divisible by \(p\text{.}\) This gives
    \begin{gather*} |\mathcal S_P| \equiv 1 \pmod{p}. \end{gather*}
    (This does not yet prove part (3) since we don’t yet know that \(\mathcal S_P\) consists of all Sylow \(p\)-subgroups.)
    We can now prove part (2): By way of contradiction, suppose \(Q\) is a \(p\)-subgroup of \(G\) such that \(Q\) is not contained in any of the subgroups belonging to \(\mathcal S_P\text{.}\) Then \(Q \cap P_i < Q\) for all \(i\) and thus every term on the right-hand side of () is divisble by \(p\text{,}\) contrary to (). The second assertion in (2) follows by taking \(Q\) to be a Sylow \(p\)-subgroup.
    This proves, in particular, that \(\mathcal S_P\) in fact does consist of all Sylow \(p\)-subgroups.
  3. Part (3) thus follows from ().
  4. For any \(P \in \operatorname{Syl}_p(G)\text{,}\) the stabilizer of \(P\) for the action of \(G\) on \(\operatorname{Syl}_p(G)\) by conjugation is \(N_G(P)\text{.}\) Since we now know the action is transitive,
    \begin{equation*} |\operatorname{Syl}_p(G)| = [G: N_G(P)]. \end{equation*}
    Moreover, since \(P \leq N_G(P)\) and \(|P| = p^e\text{,}\) it follows that \([G: N_G(P)]\mid m\text{.}\)
Sylow’s Theorems provides a systematic way of analyzing the subgroups of a finite group. It allows us to determine the number of subgroups of a given order, which is useful in many applications, such as studying the structure of solvable and simple groups. The theorems also provide information about the normalizers of subgroups, which can be used to study normal subgroups and quotient groups.

Remark 6.13.

In general, Cauchy’s Theorem can be deduced from part one of Sylow’s Theorems. For say \(p \mid |G|\text{.}\) Then there exists a Sylow \(p\)-subgroup \(P\) of \(G\text{.}\) Pick any \(x \in P\text{,}\) \(x \ne e\text{.}\) Then \(|x| = p^j\) for some \(j \geq 1\text{.}\) Then \(x^{p^{j-1}}\) has order \(p\text{.}\)
Recalling Theorem 3.24, we have the following:

Exercise 6.15. Order \(8\) of \(S_4\).

Prove that \(S_4\) has precisely three distinct subgroups of order \(8\text{,}\) all of which are isomorphic to \(D_8.\)
Solution.
Any subgroup of \(S_4\) of order \(8\) is a Sylow \(2\)-subgroup. By Sylow’s Theorems, the number of Sylow \(2\)-subgroups is either \(1\) or \(3\text{.}\) As any \(2\)-cycle or \(4\)-cycle must be contained in a Sylow \(2\)-subgroup (since such elements generate subgroups of order \(2i\text{,}\) for some \(i\)) and there are six \(2\)-cycles and six \(4\)-cycles, there has to be more than one Sylow \(2\)-subgroup. Hence, there are precisely \(3\) distinct subgroups of order \(8\text{.}\)
As all of them are conjugate, and conjugation induces an isomorphism on the group, we see that all three subgroups are isomorphic. Hence, we just need to show that \(S_4\) contains a subgroup isomorphic to \(D_8\text{.}\) Let \(X\) be the set of left cosets of the subgroup \(\{e,s\}\) of \(D_8\text{.}\) Note that \(|X| = 4\text{.}\) Let \(G\) act on \(X\) by left multiplication. This action induces a homomorphism \(\varphi : G \to P(X)\) where \(P(X)\cong S_4\) is the permutation group on \(X\text{.}\) As shown in class, the kernel of this homomorphism is the largest normal subgroup contained in \(\{e,s\}\text{,}\) which is \(\{e\}\text{.}\) Thus, \(\varphi\) is injective and the image of \(\varphi\) is a subgroup of \(S_4\) isomorphic to \(D_8\text{.}\)
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