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Let \(G\) be a finite group and \(p\) be the smallest prime factor of \(\vert G\vert\). Then every subgroup \(H\) of index \(p\) is a normal subgroup.
This proposition is a special case of this theorem.
Let \(X=G/H\). Then \(\vert X\vert =p\) and \(G\) acts on \(X\) by left multiplication, which induces a group homomorphism \(\rho: G\to S_X\). We claim that \(H\) is the kernel of \(\rho\), which implies that \(H\) is normal.
From the first isomorphism theorem, we have
\[G/\text{ker}\rho \cong \rho(G) \leq S_X,\]which means that
\[[G:\text{ker}\rho] \mid p! \implies [G:\text{ker}\rho]\mid \gcd(p!, \vert G\vert).\]\(\vert G/\text{ker}\rho\vert = [G:\text{ker}\rho]\); \(\vert S_X\vert = p!\).
Since \(p\) is the smallest prime divisor of \(\vert G\vert\), \([G:\text{ker}\rho] = 1\) or \(p\).
On the other hand, we have
\[\text{ker}\rho = \bigcap_{xH\in X} \text{Stab}_G(xH) \subset \text{Stab}_G(H) = H.\]For some \(xH\in X\) and for all \(g\in \text{Stab}_G(xH), gxH = \rho(g)xH = xH\).
然後取交集，對所有 \(xH\in X\)！
Comparing the indexes, we have
\[[G : \text{ker}\rho] = [G : H][H : \text{ker}\rho] = p[H : \text{ker}\rho].\]Together with the result \([G:\text{ker}\rho]=1\) or \(p\), we conclude that \([G:\text{ker}\rho]=p\) and \([H:\text{ker}\rho]=1\), which implies that \(H=\text{ker}\rho\). ◼
If \(\vert G\vert = pq\), where \(p>q\) are two primes, then \(G\) contains a normal subgroup of order \(p\). Moreover, if \(p\not \equiv 1\pmod q\), then \(G\) is cyclic.
Proof
Let \(p\) and \(q\) be two distinct primes with \(p>q\). Let \(G\) be a group of order \(p^kq\). Then \(G\) has a normal subgroup of order \(p^k\).
Let \(p>q>r\) be three distinct primes. Let \(G\) be a group of order \(pqr\) and \(\mathcal S_k\) be a Sylow \(k\)-subgroup of \(G\). Then
- Either \(\mathcal S_p\) or \(\mathcal S_q\) is normal.
- \(\mathcal S_p\mathcal S_q\) is a normal subgroup of \(G\).
Proof
By the Third Sylow Theorem, and prove by contradiction on the size of \(\vert G\vert\), i.e. \(pqr\).
Two subgroups \(H_1\) and \(H_2\) of a group \(G\) are called conjugate subgroups if there exists a \(g\) in \(G\) such that \(gH_1g^{-1} = H_2\). Note that this defines an equivalence relation.
Let \(H\) be a subgroup of a group \(G\). The set
\[\{g\in G\mid gHg^{-1}=H \}\]is called the normalizer of \(H\) in \(G\) denoted by \(N(H)\), which is the largest subgroup of \(G\) containing \(H\) as a normal subgroup.
Let \(X\) be the set of subgroups of \(G\) conjugate to \(H\), and let \(G\) acts on \(X\) by conjugation. For \(x=H\), we have \(\text{Stab}_G(x) = N(H)\) and
\[\vert G\vert = \vert X\vert \vert N(H)\vert ,\]by the orbit-stablizer theorem.
Let \(H\) be a \(p\)-subgroup of a finite group \(G\). Then
\[[N(H):H]\equiv [G:H] \pmod p.\]
Proof
Let \(X\) be the set of all left cosets of \(H\). Let \(H\) acts on \(X\) by left multiplication. Since \(H\) is a \(p\)-group, we have
\[\vert X\vert \equiv\vert X^H\vert \pmod p.\]Moreover, for \(gH\in X^H\),
\[(hg)H = gH,\]for all \(h\in H\), which implies that
\[g^{-1}hg \in H \implies g \in N(H).\]Therefore, \(\vert X^H\vert =[N(H):H]\) and we have
\[\vert X\vert =[G:H] \equiv [N(H):H]=\vert X^H\vert \pmod p. \tag*{$\blacksquare$}\]