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Sylow theorems can be used to find normal subgroups, and in turn determine group structures!
Let \(G\) be a group of order \(p^nm\), where \(p\) does not divide \(m\).
There exists a subgroup of order \(p^i\) for \(1<i\leq n\). Moreover, every subgroup of order \(p^i\) is normal in some subgroup of order \(p^{i+1}\) when \(i<n\).
Proof
To prove this, we construct a series of \(p\)-subgroups \(H_1 < H_2 < \cdots < H_n\), with \(\vert H_i\vert = p^i\) and \(H_i \trianglelefteq H_{i+1}\) for \(i<n\), by induction.
By Cauchy’s theorem, \(G\) has a subgroup \(H_1\) of order \(p\). Suppose that for some \(i<n\), \(H_i\) has been constructed. We then wish to construct \(H_{i+1}\).
Since \(i<n\), we have \(p\mid [G:H_i]\). By this lemma:
\[[N_G(H_i):H_i ] \equiv [G:H_i] \equiv 0\pmod p,\]which means that \(p\) divides the order of \(N_G(H_i)/H_i\). By Cauchy’s theorem, there is a subgroup \(K\) of \(N_G(H_i)/H_i\) of order \(p\).
Now consider the canonical homomorphism \(\phi: N_G(H_i)\to N_G(H_i)/H_i\) defined by \(\phi(g) = gH_i\). We can verify that \(\text{ker}(\phi) = H_i\). Suppose that \(K = \{H_i, g_1H_i,\cdots,g_{p-1}H_i \}\). We can find a subgroup \(R\) containing \(H_i\) such that \(R=\phi^{-1}(K)\). In particular, \(R = \{e, g_1,\cdots, g_{p-1} \}H_i\). Thus we have \(\vert R\vert = p^{i+1}.\)
Moreover, \(H_i\) is normal in \(N_G(H_i)\) by definition, and thus \(H_i\) is also normal in \(R\). The desired \(H_{i+1}\) is exactly \(R\). ◼
\(R = \{e, g_1,\cdots, g_{p-1} \}H_i\)，這是兩個 set 乘在一起，像 \(NH\) 一樣！
Every \(p\)-subgroup is contained in a Sylow \(p\)-subgroup.
If \(G\) is a \(p\)-group, then \(G\) admits a subgroup of any order \(d\) that divides \(\vert G\vert\).
看看這 ！
Any two Sylow \(p\)-subgroups of a finite group \(G\) are conjugate to each other.
Proof
Let \(P_1\) and \(P_2\) be two Sylow \(p\)-subgroups of \(G\). Let \(X\) be the set of all left cosets of \(P_1\) in \(G\), and let \(P_2\) act on \(X\) by left multiplication. Since \(P_2\) is a \(p\)-group, we have \(\vert X\vert\equiv \vert X^{P_2}\vert\pmod p\). On the other hand, \(\vert X \vert = [G:P_1] = m\) is not divisible by \(p\), so \(X^{P_2}\) is non-empty.
Let \(xP_1\in X^{P_2}\). Then \(gxP_1 = xP_1\) for all \(g\in P_2\), which implies that \(x^{-1}gx\in P_1\) for all \(g\in P_2\). Therefore, we have \(x^{-1}P_2x\subseteq P_1\). However \(\vert P_1\vert =\vert P_2\vert =\vert x^{-1}P_2x\vert\) , so we must have \(x^{-1}P_2x = P_1\). This shows that \(P_2\) is conjugate to \(P_1\) in \(G\), completing the proof. ◼
Conjugate 後大小不變！
Let \(H = H_1\cap H_2\cap \cdots\cap H_n\) be the intersection of all Sylow \(p\)-subgroups of \(G\). Then \(H\) is normal in \(G\).
Proof
First, we should prove a lemma:
For some \(g\in G\), \(g(H_1\cap H_2)g^{-1} = (gH_1g^{-1})\cap(gH_2g^{-1})\).
Since \(\rho_g: h \mapsto ghg^{-1}\) is an isomorphism, we have a bijective correspondence between \(H_1\) and \(gH_1g^{-1}\), and similarly, between \(H_2\) and \(gH_2g^{-1}\). Then, for all \(h\in H_1\cap H_2\), there is also a bijective correspondence between \(h\) and \(ghg^{-1}\), and thus the lemma is proved.
Now consider \(gHg^{-1}\) for some \(g \in G\). We have
\[gHg^{-1} = g\big(\bigcap_i H_i\big)g^{-1} = \bigcap_i\big(gH_ig^{-1}\big) \supseteq H,\]by the second Sylow theorem. However, it is obviois that \(gHg^{-1} \subseteq H\), and thus \(gHg^{-1}=H\). Since \(g\) is arbitrary, we have proved that \(H\) is a normal subgroup of \(G\). ◼
Let \(H = H_1\cap H_2\cap \cdots\cap H_n\) be the intersection of all Sylow \(p\)-subgroups of \(G\). If \(H\) is non-trivial, then \(G\) is not simple.
Let \(n_p\) be the number of Sylow \(p\)-subgroups. Then \(n_p\) satisfies the following two conditions:
\[\begin{align*} &n_p \equiv 1 \pmod p, \tag{1} \\ &n_p \mid m. \tag{2} \end{align*}\]
The subgroup \(H\) of order \(p^n\) is called a Sylow \(p\)-subgroup, which is exactly a maximal \(p\)-subgroup.
Proof of \(n_p \equiv 1\pmod p\)
Let \(P\) be a Sylow \(p\)-subgroup of \(G\), and let \(X\) be the set of all Sylow \(p\)-subgroups. Let \(P\) act on \(X\) by conjugation. Since \(P\) is a \(p\)-group, we have \(n_p=\vert X\vert \equiv \vert X^P\vert \pmod p\).
Let \(Q\in X^P\). Then \(P\subseteq N_G(Q)\). Since \(P\) and \(Q\) are both Sylow \(p\)-subgroups of \(N_G(Q)\) (hint: \(N_Q(G) \leq G\)) , they are conjugate in \(N_G(Q)\) by the Second Sylow Theorem. However, since \(Q\) is normal, we must have
\[P = gQg^{-1} = (Qg)g^{-1} = Q,\]for some \(g\in N_G(Q)\). This shows that \(X^P = \{P\}\), and thus \(n_p\equiv \vert X^P\vert = 1 \pmod p\).
Proof of \(n_p \mid m\)
Let \(X\) be the set of all Sylow \(p\)-subgroups, and let \(G\) act on \(X\) by conjugation. Let \(P\) be any Sylow \(p-\)subgroup. By the Second Sylow Theorem, **the \(G\)-orbit of \(P\) is the whole set \(X\). Then by the orbit-stabilizer theorem, we have
\[n_p = \vert X\vert = G_P = [G:\text{Stab}_G(P)],\]where
\[\text{Stab}_G(P) = \{g\in G\mid gPg^{-1}=P \} = N_G(P).\]Since \(P\leq N_G(P)\), we have
\[m = [G:P] = [G:N_G(P)][N_G(P):P] = n_p[N_G(P):P].\]Thus, \(n_p\mid m\). ◼
A Sylow \(p\)-subgroup is normal iff \(n_p = 1\).
對於兩質數 \(p\)、\(q\)，\(\mathcal S_p\) 和 \(\mathcal S_q\) 的交集必定只有 identity（從 element order 來想）。但對於兩個 Sylow \(p\)-subgroups，\(\mathcal{S_{p,1}}\)、\(\mathcal{S_{p,2}}\)，他們的交集不一定是 \(\{e\}\)：若 \(\vert S_{p, 1} \vert = \vert S_{p, 2} \vert = p\)，交集是 \(\{e\}\)，但若 \(\vert S_{p, 1}\vert = \vert S_{p, 2} \vert = p^k, k > 1\)，交集就不見得是 \(\{e\}\)（element order 可以是 \(p^j, j<k\)）！
假設有 \(m\) 個 Sylow \(p\)-subgroups，對於任意兩 Sylow \(p\)-subgroup，他們的交集大小必須一樣嗎？答案是否定的，看看 maximal Sylow intersection。
Check the corollaries here.
Recall: Simple groups.
A non-abelian \(p\)-group is not simple. Conversely, a non-abelian simple group is not a \(p\)-group.
Proof
By this theorem.
If \(G\) is simple, then for any non-trivial homomorphism \(\rho:G\to G'\), \(\rho\) must be injective.
Proof
Suppose \(\rho\) is not injective, which means that \(\text{ker}\rho > 1\). However, since \(\text{ker}\rho\) is normal and \(G\) is simple, we must have \(\text{ker}\rho = G\), i.e., \(\rho\) is trivial. ◼
Suppose \(G\) is a non-abelian simple group.
- For all primes \(p\mid \vert G\vert\), \(\vert G\vert \mid (n_p)!\).
- For all proper subgroups \(H\) of \(G\), \(\vert G\vert \Big\vert [G:H]!\).
Proof 1.
Since \(G\) is simple and \(G\) is not a \(p\)-group, we have \(n_p > 1\). The action of \(G\) on the set of Sylow \(p\)-subgroups by conjugation induces a group homomorphism \(\rho:G\to S_{n_p}\). By the Second Sylow Theorem, such action is transitive, which implies \(\rho\) is non-trivial. By this lemma, \(\rho\) is injective and \(\vert G\vert =\vert \rho(G)\vert\) divides \(\vert S_{n_p}\vert = (n_p)!\). ◼
Proof 2.
The action of \(G\) on \(G/H\) by left multiplication induces a group homomorphism \(\rho: G\to S_{G/H}\). Since only \(h\in H_i\) makes \(hH_i = H_i\), \(\rho\) is transitive; since \(H\) is a proper subgroup, \(\rho\) is non-trivial. Thus, \(\rho\) is injective and \(\vert G\vert = \vert \rho(G)\vert\) divides \(\vert S_{G/H}\vert= [G:H]!\).
Q1:
Show that for any non-abelian group \(G\), if \(\vert G\vert=132=2^2\cdot 3\cdot 11\), \(G\) is not a simple group.
Solution
We have \(n_2=1, 3, 11, 33\), \(n_3=1,4\), and \(n_{11} = 1, 12\). Suppose \(n_{11}=12\) and \(n_3=4\). There are already \((11-1)\cdot 12 + (3-1)\cdot 4 = 128\) elements. Suppose \(n_2=3\), for minimizing possible elements. Let \(N_2\) denote the number of elements whose order divides \(4\). Then by the inclusion-exclusion principle,
\[\begin{align*} N_2 = \vert H_1\cup H_2\cup H_3\vert &\geq \sum_i \vert H_i\vert - \sum_{\substack{i, j \\ i\not=j}}\vert H_i\cap H_j\vert \\ &\geq 4\cdot 3 - {3\choose 2}\cdot 2 \\ &= 6, \end{align*}\]where \(H_i\) are distinct Sylow \(2\)-subgroups. However, \(128 + 6=134>132\), which is a contradiction. Thus, \(G\) is not simple. ◼
\(\vert H_i\cap H_j\vert = 1\text{ or } 2\).
\(n_p\) 小的時候，inclusion-exclusion 比較容易奏效。
Q2:
Show that for any non-abelian group \(G\), if \(\vert G\vert=90\), \(G\) is not a simple group.
Solution
First note that \(90=2\cdot 3^2\cdot 5\). As a routine, we check that
\[\begin{align*} n_2&=1, 3, 5, 15 \\ n_3&=1, 10 \\ n_5&=1,6 \end{align*}\]by the third Sylow theorem. Since \(3\) is the only prime which is squared, let’s play with it. (Actually we should first suppose \(n_5=6\), and count the elements of order \(5\), trying to make a contradiction. However, this doesn’t help.)
If \(n_3=1\), game over. Suppose \(n_3=10\). For any two distinct Sylow \(3\)-subgroup \(P\) and \(Q\), there are two possibilities: \(\vert P\cap Q\vert = 1\) or \(3\).
If \(\vert P\cap Q\vert=1\), the number of elements whose order is divisible by \(3\) is \(10 \cdot (9 - 1) = 80\). Add this with the number of order \(5\) elements, we have \(6\cdot(5-1) + 80 = 104 > 90\), a contradiction. Thus, we must have some \(P, Q\) such that \(\vert P\cap Q\vert = 3\).
In such case, by the second isomorphism theorem,
\[\vert PQ\vert = {\vert P\vert \vert Q\vert \over \vert P\cap Q\vert } = 27.\]Then, by the first Sylow theorem, we know that \(P\cap Q\) is normal in \(P\) and in \(Q\)(or use this theorem), which implies that
\[PQ \subseteq N_G(P\cap Q) \text{ and } P\leq N_G(P\cap Q).\]\(P\) and \(Q\) are assumed to be not normal, and thus \(PQ\) is not necessarily a subgroup!
Hence, we have gathered some information regarding the order of \(N_G(P\cap Q)\):
the only choices are \(45\) and \(90\). If \(\vert N_G(P\cap Q)\vert = 90\), we have \(P\cap Q\) as a non-trivial proper normal subgroup of \(G\). If \(\vert N_G(P\cap Q)\vert = 45\), since \([G:N_G(P\cap Q)] = 2\), we have \(N_G(P\cap Q)\) as a normal subgroup of \(G\). In conclusion, \(G\) must not be simple. ◼
Reference: StackExchange
For \(\vert G\vert = 112\), check this. (Check all the answers, and comments.)
Q3:
Show that for any non-abelian group \(G\), if \(\vert G\vert=120\), \(G\) is not a simple group.
Solution
Assume \(G\) is simple. Then we have \(n_5=6\). This means (check the proof here) that \(\rho: G\to S_6\) is injective. Moreover, since \(2\) is the minimum prime factor of \(\vert G\vert=120\), \(G\) has no subgroup of index \(2\), and neither does \(\rho(G)\). This implies that \(\rho(G)\) is composed of only even cycles (Otherwise, half of \(\rho(G)\) is even and the other half is odd, which means that \(\rho(G)\) has a subgroup of index \(2\).), i.e., \(\rho(G)\le A_6\).
Now, let \(A_6\) acts on \(A_6/\rho(G)\). Since this action is transitive (multiply by some inverse), \(\rho': A_6 \to S_{\vert A_6/\rho(G)\vert} \cong S_3\) is non-trivial, i.e., \(\text{ker}(\rho') \neq A_6\). Moreover, since \(A_6\) is simple (\(A_n\) is simple except \(n=4\).), we must have \(\text{ker}(\rho') = \{e\}\). However, \(\vert A_6\vert = 360 > 6 = \vert S_3\vert\), which is a contradiction. Therefore, \(G\) is not simple. ◼
Reference: Aryaman