This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
\(p\)-Groups
In this post, \(p\) always denotes a prime.
Definition 1 A finite group \(G\) is called a \(p\)-group if its order is a power of \(p\).
Then it is obvious that any subgroup and quotient group of a \(p\)-group is again a \(p\)-group. Moreover, the following holds.
Lemma 2 Suppose a \(p\)-group \(G\) acts on a finite set \(E\), and consider the set of fixed points of this action
\[E^G=\{x\in E\mid g\cdot x=x\text{ for all $g\in G$}\}\]Then
\[\lvert E^G\rvert\equiv\lvert E\rvert\pmod{p}\]holds.
Proof
In other words, we must show that the size of the set \(E\setminus E^G\) is a multiple of \(p\). However, \(E\setminus E^G\) is a union of (disjoint) \(G\)-orbits each of size greater than \(1\), and the size of each such orbit is a power of \(p\) by [Algebraic Structures] §Group Actions, ⁋Theorem 14, so this holds.
In particular, if we consider the case where \(G\) acts on \(E=G\) by inner automorphisms, then \(E^G\) is precisely the center of \(G\), so by Lemma 2 we know that the center \(C(G)\) of a \(p\)-group \(G\) is nontrivial.
Theorem 3 For a \(p\)-group \(G\) of order \(p^r\), there exists a series of subgroups of \(G\)
\[G=G_1\supset G_2\supset\cdots G_{n+1}=\{e\}\]such that \([G, G_k]\subseteq G_{k+1}\) holds for every \(k\), and each \(G_k/G_{k+1}\) is a cyclic group of order \(p\).
Proof
We prove this by induction on the order of \(G\). First, the case \(G=\{e\}\) is trivial. Now assume the given statement holds for all \(p\)-groups of order smaller than \(p^r\), and let us prove the case where \(\lvert G\rvert=p^r\). From the preceding argument we have \(C(G)\neq\{e\}\), so we can choose an element \(x\in C(G)\) whose order is \(p^s\) (\(1\leq s\leq r\)).
Now consider the subgroup \(H\) of \(C(G)\) generated by the element \(x^{p^{s-1}}\). Then \(G'=G/H\) is a \(p\)-group of order \(p^{r-1}\), so by the inductive hypothesis there exists a series of subgroups satisfying the given condition, and taking the inverse image of this series under the canonical projection \(p: G \rightarrow G'\) gives the desired series.
Therefore, by the equivalence of the first and second conditions in §Series of Groups, ⁋Proposition 7, we see that every \(p\)-group is always nilpotent.
On the other hand, by §Series of Groups, ⁋Proposition 8 we obtain the following.
Proposition 4 Fix a \(p\)-group \(G\) and a subgroup \(H\subsetneq G\).
- The normalizer \(N_G(H)\) of \(H\) in \(G\) satisfies \(N_G(H)\subsetneq G\).
- There exists a normal subgroup \(N\) of \(G\) of index \(p\) containing \(H\).
Therefore, any subgroup of index \(p\) of a \(p\)-group \(G\) is always normal.
Sylow Theorems
We now examine the \(p\)-subgroups of a general group. Of particular interest among them is the following.
Definition 5 A Sylow \(p\)-subgroup of a finite group \(G\) is a subgroup \(P\) of \(G\) satisfying the following two conditions.
- \(P\) is a \(p\)-group.
- \([G:P]\) is not divisible by \(p\).
We denote the set of Sylow \(p\)-subgroups of \(G\) by \(\Syl_p(G)\).
That is, when the order of \(G\) is given as \(p^r m\) (\(p\not\mid p\)), a Sylow \(p\)-group \(P\) is exactly a subgroup of \(G\) of order \(p^r\). For the rest of this post, we assume that the order \(n=p^rm\) of \(G\) always satisfies \(p\not\mid m\).
The Sylow theorems are theorems about the Sylow \(p\)-subgroups of an arbitrary finite group, and they help in classifying finite groups. The first result concerns the existence of a Sylow \(p\)-subgroup, and for this we first need the following lemma.
Lemma 6 Let \(n=p^rm\) with \(p\nmid m\). Then
\[\binom{n}{p^r}\not\equiv 0\pmod{p}\]holds.
Proof
Consider a group \(G\) of order \(p^r\) and a set \(S\) of size \(m\). Then the set \(G\times S\) is a set of size \(n\), and defining the set \(E\) as the set of subsets of \(G\times S\) of size \(p^r\) we have
\[\lvert E\rvert=\binom{n}{p^r}\]If \(G\) acts on \(G\times S\) by the formula
\[g \cdot (x, s) = (g x, s) \quad (g, x \in G,\; s \in S)\]then by applying this action to the elements of each subset in \(E\) (i.e., the subsets of \(G\times S\) of size \(p^r\)), we obtain an action of \(G\) on \(E\). The set of fixed points \(E^G\) of this action consists entirely of sets of the form
\[G \times \{s\},\qquad s\in S\]so \(\lvert E^G\rvert=m\), and now by Lemma 2 we have
\[\binom{n}{p^r} = \text{Card}(E) \equiv \text{Card}(E^G) = m \not\equiv 0 \pmod{p}\]Then the first of the Sylow theorems is about the existence of a Sylow \(p\)-subgroup.
Theorem 7 \(G\) has a Sylow \(p\)-subgroup.
Proof
Let \(E\) be the set of subsets of \(G\) with exactly \(p^r\) elements. Then, by Lemma 6
\[\lvert E\rvert = \binom{n}{p^r}\not\equiv 0\pmod{p}\]Now consider the left translation action of \(G\) on itself
\[L_g:G \rightarrow G;\qquad x\mapsto gx\]and regard this as an action on \(E\) in the same way as in the proof of Lemma 6. Then from the assumption that \(\lvert E\rvert\not\equiv 0\pmod{p}\), there exists an orbit \(O\) whose size is not a multiple of \(p\). Now let \(X\) be an element of \(O\), and let the stabilizer of \(X\) be \(\Stab(\{X\})=\Stab(X)\). Then \(\Stab(X)\) is a subgroup of \(G\), and ([Algebraic Structures] §Group Actions, ⁋Corollary 8) this is the subgroup we want.
To show this, first from [Algebraic Structures] §Group Actions, ⁋Theorem 14 we have
\[\lvert O\rvert=\lvert G\cdot X\rvert=[G:\Stab(X)]=\frac{\lvert G\rvert}{\lvert\Stab(X)\rvert}\not\equiv 0\pmod{p}\]so \(p^r\) divides \(\lvert \Stab(X)\rvert\).
On the other hand, \(\Stab(X)\) is the set of elements \(g\in G\) satisfying \(gX = X\), and thus for any element \(x \in X\) we have
\[\Stab(X) \subseteq X x^{-1}\]so
\[\lvert \Stab(X)\rvert\leq\lvert Xx^{-1}\rvert=\lvert X\rvert=p^r\]must hold. From this we see that \(\lvert\Stab(X)\rvert=p^r\).
From this we know that if the order of an arbitrary finite group \(G\) is divisible by \(p\), then \(G\) has an element of order \(p\).
We say that two subgroups \(H_1, H_2\) of \(G\) are conjugate if there exists \(\rho\in\Inn(G)\) such that \(\rho(H_1)=H_2\).
Theorem 8 The following hold.
- The Sylow \(p\)-subgroups of \(G\) are conjugate to one another, and their number is \(1\) mod \(p\).
- Every \(p\)-subgroup of \(G\) is contained in some Sylow \(p\)-subgroup.
Proof
Let \(P\) be a Sylow \(p\)-subgroup of \(G\), and let \(H\) be a \(p\)-subgroup of \(G\). Considering the left translation action of \(H\) on the set \(E = G/P\), by Lemma 6 we have \(\lvert E^H\rvert\neq 0\), so there exists \(x\in G/P\) such that \(Hx=x\). Now choose a representative \(g\in G\) of the element \(x\in G/P\). Then for any \(h \in H\) we have \(h(gP) = gP\), so \(g^{-1} h g \in P\). Therefore \(H \subseteq gPg^{-1}\), and this proves the second claim.
Now suppose \(H\) is a Sylow \(p\)-subgroup. Then
\[\lvert H \rvert = \lvert P \rvert = \lvert gPg^{-1} \rvert\]so the above inclusion becomes \(H = gPg^{-1}\), proving the first part of the first claim.
To prove the latter part of the first claim, let \(G\) act on \(\Syl_p(G)\) by inner automorphisms. Then from the preceding argument, any \(P \in \Syl_p(G)\) is a fixed point of this action, and we show that this is the unique fixed point.
Suppose for contradiction that there exists another fixed point \(Q \in \Syl_p(G)\). \(Q\) is a Sylow \(p\)-subgroup of \(G\) normalized by \(P\). That is, \(P\subseteq N_G(Q)\). Now \(P\) and \(Q\) are both Sylow \(p\)-subgroups of \(N_G(Q)\), so by the preceding argument there exists an appropriate \(n \in N_G(Q)\) such that
\[P = nQn^{-1} = Q\]Therefore, from Lemma 6 we know that \(\lvert \Syl_p(G) \rvert = \lvert \Syl_p(G)^P \rvert \equiv 1 \pmod{p}\).
Corollary 9 Consider \(P\in\Syl_p(G)\) and its normalizer \(N_G(P)\). For a subgroup \(M\) of \(G\) containing \(N_G(P)\), the normalizer \(N_G(M)\) of \(M\) in \(G\) equals \(M\).
Proof
Choose \(g\in G\) satisfying \(M=gMg^{-1}\). Then \(gPg^{-1}\) is a Sylow \(p\)-subgroup of \(M\). Hence there exists an appropriate \(h \in M\) such that \(gPg^{-1} = hPh^{-1}\). Now \(h^{-1}g \in N\), and therefore \(g \in hN \subset M\).
Corollary 10 Fix a group homomorphism \(f: G_1 \to G_2\) between finite groups. For a Sylow \(p\)-subgroup \(P_1\) of \(G_1\), there exists a Sylow \(p\)-subgroup \(P_2\) of \(G_2\) containing \(f(P_1)\).
Proof
Apply the second result of Theorem 8 to the subgroup \(f(P_1)\) of \(G_2\).
Corollary 11
- Let \(H\) be a subgroup of \(G\). For a Sylow \(p\)-subgroup \(P\) of \(H\), there exists a Sylow \(p\)-subgroup \(Q\) of \(G\) such that \(P = Q \cap H\).
- Conversely, let \(Q\) be a Sylow \(p\)-subgroup of \(G\) and let \(H\) be a normal subgroup of \(G\). Then \(Q \cap H\) is a Sylow \(p\)-subgroup of \(H\).
Proof
- The \(p\)-group \(P\) is contained in a Sylow \(p\)-subgroup \(Q\) of \(G\). On the other hand, \(Q \cap H\) is a \(p\)-subgroup of \(H\) containing \(P\), hence \(P = Q \cap H\).
- Let \(P'\) be a Sylow \(p\)-subgroup of \(H\). Then there exists an appropriate \(g \in G\) such that \(gP'g^{-1} \subset Q\). Since \(H\) is a normal subgroup, \(P = gP'g^{-1}\) is again contained in \(H\), and thus \(P\) is contained in \(Q\cap H\). Now \(Q \cap H\) is a \(p\)-subgroup of \(H\), and \(P\) is a Sylow \(p\)-subgroup, so \(P = Q \cap H\).
Corollary 12 Let \(N\) be a normal subgroup of \(G\). Then the image in \(G/N\) of a Sylow \(p\)-subgroup of \(G\) is a Sylow \(p\)-subgroup of \(G/N\), and moreover every Sylow \(p\)-subgroup of \(G/N\) is obtained in this way.
Proof
Fix \(P\in \Syl_p(G)\), let \(G' = G/N\), and let \(P'\) be the image of \(P\) in \(G'\).
Considering the left translation action of \(G\) on \(G'/P'\), this is a transitive action, so the orbit of \(G\) is \(G'/P'\) itself. Now by [Algebraic Structures] §Group Actions, ⁋Theorem 14 we have
\[\lvert G'/P'\rvert=[G:\Stab(G'/P')]\]But by definition \(\Stab(G'/P')\) contains \(P\), so \([G:\Stab(G'/P')]\) is not divisible by \(p\), and hence \([G':P']\) is also not divisible by \(p\). On the other hand, since \(P'\) is a \(p\)-group, by definition \(P'\) is a Sylow \(p\)-subgroup of \(G'\).
For the converse, consider another Sylow \(p\)-subgroup \(Q'\) of \(G'\). Then \(Q' = g'P'g'^{-1}\) for some \(g' \in G'\), and taking a representative \(g \in G\) of \(g'\), the image of \(gPg^{-1}\) is \(Q'\).
Applications of the Sylow Theorems
As mentioned above, the Sylow theorems are usefully applied to the classification of finite groups. To this end, let us examine Theorem 8 a little more closely. Let \(n_p\) denote the size of \(\Syl_p(G)\). Then by the latter part of the first result of Theorem 8 we have \(n_p\equiv 1\pmod{p}\). On the other hand, the first part of the first result of Theorem 8 shows that \(G\) acts transitively on \(\Syl_p(G)\), so by [Algebraic Structures] §Group Actions, ⁋Theorem 14 we obtain
\[n_p=\lvert \Syl_p(G)\rvert=[G:\Stab(P)],\qquad P\in\Syl_p(G)\]and in particular \(n_p\) must divide \(\lvert G\rvert\); moreover, as we saw above, \(n_p\) does not divide \(p^r\), so \(n_p\) must divide \(m\).
Example 13 Let us classify a finite group \(G\) of order \(15\).
\[\lvert G\rvert = 15 = 3\times 5\]First consider the Sylow 3-subgroups. Then by Theorem 8 the number \(n_3\) of Sylow 3-subgroups satisfies the following two conditions:
- \(n_3\equiv 1\pmod{3}\),
- \(n_3\) divides \(5\).
The only \(n_3\) satisfying these two conditions is \(1\), and by the result of Theorem 8 this means that the (unique) Sylow \(3\)-subgroup \(P_3\) of \(G\) is a normal subgroup.
Similarly, let us consider the Sylow 5-subgroups. By the Sylow theorems, the number \(n_5\) of Sylow 5-subgroups satisfies the following conditions:
- \(n_5\equiv 1\pmod{5}\),
- \(n_5\) divides \(3\).
Likewise, the only \(n_5\) satisfying these two conditions is also \(1\), so the Sylow 5-subgroup also exists uniquely and is a normal subgroup. Let us denote it by \(P_5\).
Now \(P_3\cap P_5\) is a subgroup of both \(P_3\) and \(P_5\), so its order must be \(1\), and therefore \(P_3\cap P_5=\{e\}\). Now considering the subgroup \(P_3P_5\) of \(G\), from [Algebraic Structures] §Isomorphism Theorems, ⁋Theorem 5 we have
\[\frac{P_3P_5}{P_3}\cong P_5/\{e\}\implies \lvert P_3P_5\rvert=\lvert P_3\rvert\lvert P_5\rvert=15=\lvert G\rvert\]so ultimately we must have \(G\cong \mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}\).
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