Lagrange and Cauchy
Lagrange's theorem (Tag T-GJNT5)
Let \(G\) be a finite group and \(H \leq G\) a subgroup. Then \(\abs H\) divides \(\abs G\), and \([G:H] = \abs G/ \abs H\) [@DF04].
The left cosets of \(H\) partition \(G\) (PR-VUKHO), and each coset \(gH\) has \(\abs H\) elements because \(h\mapsto gh\) is a bijection \(H\to gH\). Hence \(\abs G=[G:H]\,\abs H\).
(of Lagrange's theorem)
Let \(N \da [G:H]\) and write \(G/H = \ts{g_1 H, \ldots, g_N H}\). Two left cosets are equal or disjoint, and \(x\mapsto g_kx\) is a bijection \(H\to g_kH\), so \[ G = \disjoint_{k=1}^N g_k H \implies \size G = \sum_{k=1}^N \size{g_k H} = \sum_{k=1}^N \size H = N \size H. \] Hence \(\size G = [G:H]\,\size H\), and both \(\size H\) and \([G:H]\) divide \(\size G\).
Coset counting formula $\# G = [G:H]\,\#H$ (Tag C-HWX2P)
Let \(G\) be a finite group, let \(H\leq G\) be a subgroup, and let \(G/H\) be the set of left cosets of \(H\) in \(G\), so that \([G:H] = \size (G/H)\) is the index of \(H\). Then \[ \size G = [G:H]\, \size H, \qquad\text{equivalently}\qquad [G:H] = \frac{\size G}{\size H}. \]
For a finite group \(G\) and \(g\in G\), the order of \(g\) divides \(\size G\), and \(g^{\size G} = e\).
(The converse of Lagrange's theorem is false)
A finite group \(G\) need not have a subgroup of order \(n\) for every \(n \divides \size G\).
The group \(A_5\) has order \(60\) and no subgroup of order \(30\): such a subgroup would have index \(2\), hence be normal, and \(A_5\) is simple.
The group \(A_4\) has order \(12\) and no subgroup of order \(6\). Suppose \(H\le A_4\) has order \(6\). The \(3\)-cycles generate \(A_4\), so some \(3\)-cycle \(x\) satisfies \(x\notin H\). Since \([A_4 : H] = 2\), two of the cosets \(H, xH, x^2H\) coincide. But \(H = xH\) gives \(x\in H\); \(x^2 H = H\) gives \(\inverseof{x} = x^2 \in H\), hence \(x\in H\); and \(xH = x^2H\) gives \(x = \inverseof{x} x^2 \in H\). \(\contradiction\)
Cauchy's theorem (Tag T-3KCD6)
Let \(G\) be a finite group and \(p\) a prime dividing \(\abs{G}\). Then \(G\) has an element of order \(p\), and hence a subgroup of order \(p\).
By Lagrange's theorem, the order of every element of \(G\) divides \(\abs G\); conversely, Cauchy’s theorem gives an element of order \(p\) for every prime \(p\) dividing \(\abs G\). Sylow's first theorem strengthens it: if \(p^a\) is the largest power of \(p\) dividing \(\abs G\), then \(G\) has a subgroup of order \(p^a\).
Let \(p\) be a prime dividing \(\size G\), and let \(X = \ts{(g_1,\ldots,g_p)\in G^p \st g_1g_2\cdots g_p = e}\). Choosing \(g_1,\ldots,g_{p-1}\) freely determines \(g_p\), so \(\size X = \size G^{p-1}\), which is divisible by \(p\). The group \(\ZZ/p\) acts on \(X\) by cyclic rotation of the coordinates, since \(g_1g_2\cdots g_p = e\) implies \(g_2\cdots g_pg_1 = \inverseof{g_1} e\, g_1 = e\). Each orbit has size \(1\) or \(p\), and the orbits of size \(1\) are the tuples \((g,\ldots,g)\) with \(g^p = e\). Hence the number of \(g\in G\) with \(g^p=e\) is congruent to \(\size X\equiv 0 \pmod p\); it is at least \(1\), because \(g=e\) qualifies, so it is at least \(p\), and some \(g\neq e\) has order \(p\).
Actions
Group action (Tag D-3T6O2)
Let \(G\) be a group with identity \(e\), and let \(X\) be a set. An action of \(G\) on \(X\) is a map \[ G \times X \to X, \qquad (g,x) \mapsto g\cdot x, \] such that for all \(g,h\in G\) and \(x\in X\):
\(e\cdot x = x\);
\(g\cdot (h\cdot x) = (gh)\cdot x\).
Let \(G\) be a group, let \(X\) be a set, and let \(\Aut(X)\) be the group of bijections \(X\to X\) under composition. Actions of \(G\) on \(X\) correspond bijectively to group homomorphisms \(\phi\colon G\to\Aut(X)\): an action corresponds to \(\phi(g)(x) \coloneqq g\cdot x\), and a homomorphism \(\phi\) corresponds to the action \(g\cdot x \coloneqq \phi(g)(x)\).
Lying in the same orbit is an equivalence relation, so the orbits partition \(X\), and \(G\) acts transitively on each orbit. A point \(x\) is fixed if and only if \(\Orb(x) = \ts x\), equivalently \(\Stab_G(x) = G\). The notation is listed on Notation.
For an action \(\psi\colon G\to\Aut_\Set(X)\), the kernel is the intersection of the stabilizers: \[ \ker \psi = \Intersect_{x\in X} G_x. \]
Transitive group action (Tag D-KGGWK)
Let \(G\) be a group acting on a set \(X\). The action is transitive if for all \(x, y\in X\) there exists \(g\in G\) such that \(g\cdot x = y\).
For nonempty \(X\), the action is transitive if and only if \(X\) is a single orbit: \(G\cdot x=X\) for some, equivalently every, \(x\in X\).
Transitive group action (Tag FD-KSKDG)
Let a group \(G\) act on a set \(X\). The action is transitive if for every pair \(x, y\in X\) there exists \(g\in G\) with \(g\cdot x = y\).
Stabilizer $G_x$ of a point (Tag FD-IGEOR)
Let a group \(G\) with identity \(e\) act on a set \(X\), and let \(x\in X\). The stabilizer of \(x\) is \[ G_x \coloneqq \theset{g\in G \st g\cdot x = x}. \]
\(G_x\) is a subgroup of \(G\): \(e\cdot x=x\); if \(g,h\in G_x\), then \((gh)\cdot x=g\cdot(h\cdot x)=x\); and if \(g\in G_x\), then \(g^{-1}\cdot x=g^{-1}\cdot(g\cdot x)=x\).
The orbit-stabilizer theorem
Orbit-stabilizer bijection for a transitive action (Tag PR-GSDKO)
Let a group \(G\) act transitively on a set \(X\), and let \(x \in X\) with stabilizer \(G_x\). Then \[ \begin{aligned} \Phi\colon G/G_{x} &\to X \\ gG_{x} &\mapsto g \cdot x \end{aligned} \] is a well-defined bijection from the set of left cosets of \(G_x\) to \(X\), and \(\Phi(h \cdot gG_x) = h \cdot \Phi(gG_x)\) for all \(g, h \in G\).
(of orbit-stabilizer)
Define \(\Phi\colon G/G_x\to\Orb(x)\) by \(\Phi(gG_x) = g\actson x\).
\(\Phi\) is well defined: if \(gG_x = hG_x\), then \(\inverseof{g} h \in G_x\), so \(h\actson x = g\actson\big((\inverseof{g} h)\actson x\big) = g\actson x\).
\(\Phi\) is injective: if \(g\actson x=h\actson x\), then \(\inverseof{g} h \actson x = x\), so \(\inverseof{g} h \in G_x\) and \(gG_x = hG_x\).
\(\Phi\) is surjective, since every element of \(\Orb(x)\) has the form \(g\actson x\).
Stabilizers of points in the same orbit are conjugate (Tag PR-KGHJ2)
Let a group \(G\) act on a set \(X\), and let \(x_0, x_1 \in X\) lie in the same orbit, say \(x_1 = g \cdot x_0\) with \(g \in G\). Then the stabilizers satisfy \(G_{x_1} = g\, G_{x_0}\, \inverseof{g}\). In particular, if \(G\) acts transitively on \(X\), all stabilizers are conjugate.
(that stabilizers along an orbit are conjugate)
Let \(x\in X\) and \(y\in \Orb(x)\), choose \(g\in G\) with \(g\actson x=y\), and write \(H_x \da \Stab(x)\) and \(H_y\da \Stab(y)\). Then \[ \begin{aligned} h\in H_x &\iff hx = x \\ &\iff h\inverseof{g} y = \inverseof{g} y \\ &\iff gh\inverseof{g} y = y \\ &\iff gh\inverseof{g} \in H_y \\ &\iff h\in \inverseof{g} H_y g, \end{aligned} \] so \(H_x = \inverseof{g} H_y g\).
Orbit-stabilizer theorem (Tag T-QYDVH)
Let a group \(G\) act on a set \(X\), and let \(x\in X\) have orbit \(Gx\) and stabilizer \(G_x\). Then \[ \size{Gx} = [G: G_x], \] and if \(G\) is finite, then \(\size{Gx} = \size{G} / \size{G_x}\).
The map \(G/G_x\to Gx\), \(gG_x\mapsto gx\), is a well-defined bijection of sets, which gives the equality \(\size{Gx} = [G:G_x]\).
Fixed points and nontrivial orbits
For an action \(\phi\) of a finite group \(G\) on a finite set \(X\), let \(x_1,\ldots,x_r\) be representatives of the orbits of size greater than \(1\). Then \[ \size X = \size{\Fix(\phi)} + \sum_{i=1}^r [G:\Stab(x_i)], \] and each \([G:\Stab(x_i)]\) is a divisor of \(\size G\) greater than \(1\).
The orbits partition \(X\), the orbits of size \(1\) are the fixed points, and \(\size{\Orb(x_i)} = [G:\Stab(x_i)]\) by orbit-stabilizer.
The class equation is this formula for the conjugation action of \(G\) on itself; see The class equation.
Four actions
(Left translation on $G$)
\(G\) acts on itself by \(\phi\colon g \mapsto (h\mapsto gh)\).
\(\Orb(x) = G\), so the action is transitive.
\(\Stab(x) = \ts e\), and \(\Fix(\phi) = \emptyset\) unless \(G\) is trivial.
The kernel is trivial, so \(G\) embeds in the symmetric group on the set \(G\) (Cayley’s theorem).
Orbit-stabilizer gives the bijection \(G/\ts e \to G\).
(Conjugation on $G$: centers and centralizers)
\(G\) acts on itself by \(g\actson x = gx\inverseof{g}\).
\(\Orb(x) = [x]\) is the conjugacy class of \(x\). The orbit of \(e\) is \(\ts e\), so the action is transitive only when \(G\) is trivial; every orbit is a singleton if and only if \(G\) is abelian.
\(\Fix(\phi) = Z(G)\), the center.
\(\Stab(x) = C_G(x)\), the centralizer of \(x\).
The kernel is \(Z(G)\).
Orbit-stabilizer gives a bijection \(G/C_G(x)\to[x]\), so the size of a conjugacy class is the index of the centralizer.
(Conjugation on subgroups: normalizers)
\(G\) acts on \(\ts{H \st H\leq G}\) by conjugation.
\(\Orb(H) = \ts{gH\inverseof{g} \st g\in G}\) is the set of conjugates of \(H\).
\(\Fix(\phi)\) is the set of normal subgroups of \(G\).
\(\Stab(H) = N_G(H)\) is the normalizer of \(H\).
The kernel is \(\Intersect_{H\leq G} N_G(H)\).
Orbit-stabilizer gives the number of conjugates of \(H\): \[ \size{\ts{ gHg ^{-1} \st g \in G } } = [G: N_G(H)]. \]
(Left translation on cosets: the normal core)
For a proper subgroup \(H < G\), \(G\) acts on \(G/H\) by left translation.
\(\Orb(eH) = G/H\), so the action is transitive.
\(\Stab(xH) = xH\inverseof{x}\), since \[ \begin{aligned} \Stab(xH) &= \ts{g\in G\st gxH = xH} \\ &= \ts{g\in G \st \inverseof{x} g x\in H} \\ &= xH\inverseof{x}. \end{aligned} \]
\(\Fix(\phi) = \emptyset\), because the action is transitive and \(\size{G/H} > 1\).
The kernel is \(\Intersect_{g\in G} gH\inverseof{g}\), the normal core of \(H\), which is the largest normal subgroup of \(G\) contained in \(H\).
This action gives the index bounds on Show $G$ is not simple.
Exercises
Poincaré's theorem on finite-index subgroups (Tag E-M6XGF)
Prove Poincaré’s theorem: if \(H\le G\) has finite index \(n\), then there is \(N\trianglelefteq G\) with \(N\subseteq H\) and \[ [G:N]\mid n!. \]
Solution
Let \(G\) act by left multiplication on the \(n\) left cosets \(G/H\). This gives a homomorphism \[ \rho:G\longrightarrow S_n. \] Let \[ N=\ker\rho. \] Then \(N\trianglelefteq G\). Moreover, \[ g\in N \iff gxH=xH\text{ for every }x\in G \iff x^{-1}gx\in H\text{ for every }x\in G, \] so \[ N=\bigcap_{x\in G}xHx^{-1}\subseteq H. \] By the first isomorphism theorem, \[ G/N\cong\operatorname{im}\rho\le S_n. \] Hence, by Lagrange, \[ [G:N]=|G/N|\mid |S_n|=n!. \]