Definitions
For a finite group \(G\) and \(g\in G\), the order of \(g\) divides \(\size G\), by Lagrange’s theorem applied to the cyclic subgroup \(\langle g\rangle\).
Injective maps are the maps with a left inverse (Tag FD-JI2RH)
Let \(X\) be a nonempty set and \(f\colon X\to Y\) a map of sets. Then \(f\) is injective, that is, \(f(x) = f(x')\) implies \(x=x'\) for all \(x,x'\in X\), if and only if \(f\) has a left inverse: a map \(g\colon Y\to X\) with \(g(f(x))=x\) for every \(x\in X\).
If \(g\) is a left inverse and \(f(x)=f(x')\), then \(x=g(f(x))=g(f(x'))=x'\). Conversely, suppose \(f\) is injective and fix \(x_0\in X\). Define \(g(y)\) to be the unique \(x\in X\) with \(f(x)=y\) when \(y\in f(X)\), and \(g(y)\coloneqq x_0\) otherwise; then \(g(f(x))=x\) for every \(x\in X\).
The hypothesis \(X\neq\emptyset\) is needed: for a nonempty set \(Y\), the empty map \(\emptyset\to Y\) is injective but has no left inverse, since there is no map \(Y\to\emptyset\).
Surjective maps are the maps with a right inverse (Tag FD-BWVRA)
Let \(f\colon X\to Y\) be a map of sets. Then \(f\) is surjective, that is, for every \(y\in Y\) there exists \(x\in X\) with \(f(x)=y\), if and only if \(f\) has a right inverse: a map \(g\colon Y\to X\) with \(f(g(y))=y\) for every \(y\in Y\).
If \(g\) is a right inverse, then each \(y\in Y\) equals \(f(g(y))\), so \(f\) is surjective. Conversely, if \(f\) is surjective, then each fiber \(f^{-1}(y)\) is nonempty, and the axiom of choice gives a map \(g\colon Y\to X\) with \(g(y)\in f^{-1}(y)\) for every \(y\); then \(f(g(y))=y\).
Order of a group element (Tag D-ZTPHF)
Let \(G\) be a group with identity \(e\) and \(g\in G\). If \(g^n=e\) for some integer \(n\geq 1\), the order of \(g\), written \(o(g)\), is the smallest such \(n\). If no such \(n\) exists, \(g\) has infinite order.
Element orders divide the order of a finite group (Tag E-DKMHL)
Show that the order of any element in a finite group divides the order of the group.
Solution
Let \(G\) be finite and let \(g\in G\).
<1>1. The cyclic subgroup \(\langle g\rangle\) has order equal to the order of \(g\).
By definition, the order \(|g|\) is the least positive integer \(n\) such that \(g^n=e\). Then \[ \langle g\rangle=\{e,g,g^2,\ldots,g^{n-1}\} \] has exactly \(n=|g|\) elements.
<1>2. Lagrange’s theorem gives \[ |G|=[G:\langle g\rangle]\,|\langle g\rangle|. \]
The left cosets of the subgroup \(\langle g\rangle\) partition the finite group \(G\), and each coset has \(|\langle g\rangle|\) elements. Their number is the index \([G:\langle g\rangle]\).
<1>3. Therefore \(|g|\) divides \(|G|\).
By <1>1 and <1>2, \[ |G|=[G:\langle g\rangle]\,|g|, \] and the index is a positive integer.
Group presentation (Tag D-V6UJ7)
Let \(S\) be a set, let \(F[S]\) be the free group on \(S\), and let \(R\subseteq F[S]\) be a set of words in \(S\). The group presented by generators \(S\) and relations \(R\) is \[ \gens{S \st R} \coloneqq F[S] / \cl_n(R), \] where \(\cl_n(R)\) is the normal closure of \(R\) in \(F[S]\), the smallest normal subgroup of \(F[S]\) containing \(R\). A presentation of a group \(G\) is an isomorphism \(G\cong\gens{S \st R}\).
Subgroups
A nonempty subset \(H\subseteq G\) is a subgroup if and only if \(xy^{-1}\in H\) for all \(x,y\in H\). The subgroup generated by a subset \(S\) is the intersection of all subgroups containing \(S\), so a property that holds for every subgroup containing \(S\) holds for \(\langle S\rangle\). The commutator \([x,y]\) is trivial if and only if \(x\) and \(y\) commute, and the commutators of \(G\) generate the commutator subgroup \([G,G]\), the first term of the derived series.
Subgroup (Tag D-IQ4OX)
Let \(G\) be a group with identity \(e\). A subset \(H\subseteq G\) is a subgroup of \(G\) if
\(gh\in H\) for all \(g,h\in H\);
\(e\in H\);
\(g^{-1} \in H\) for all \(g\in H\).
Subgroup generated by a subset (Tag D-D36IY)
Let \(G\) be a group and let \(H\subseteq G\) be a subset. The subgroup generated by \(H\) is the intersection of all subgroups of \(G\) containing \(H\): \[ \gens{H} \coloneqq \bigcap \theset{M \suchthat H\subseteq M \leq G}. \]
Let \(G\) be a group and let \(H\subseteq G\) be a subset. Then \(\gens{H}\) is the smallest subgroup of \(G\) containing \(H\), and \[ \gens{H} = \theset{ h_1^{\varepsilon_1} \cdots h_n^{\varepsilon_n} \suchthat n\geq 0,\ h_i \in H,\ \varepsilon_i\in\theset{1,-1}}. \]
Commutator subgroup (Tag D-BQ4BQ)
Let \(G\) be a group. For \(a, b\in G\), the commutator of \(a\) and \(b\) is \([a,b] \coloneqq aba^{-1}b^{-1}\). The commutator subgroup of \(G\) is the subgroup generated by all commutators: \[ [G, G] \coloneqq \gens{ aba^{-1} b^{-1} \st a, b\in G }. \]
Let \(G\) be a group. Then \([G,G]\) is a normal subgroup of \(G\) and \(G/[G,G]\) is abelian. If \(N\normal G\) and \(G/N\) is abelian, then \([G, G]\leq N\).
An element of \([G, G]\) is a product of commutators, and need not itself be a commutator.
One-step subgroup test (Tag PR-U5CHT)
Let \(G\) be a group and \(H \subseteq G\) a nonempty subset. If \(ab^{-1}\in H\) for all \(a,b\in H\), then \(H\) is a subgroup of \(G\).
Choose \(a\in H\); then \(1=aa^{-1}\in H\). For \(b\in H\), \(b^{-1}=1\cdot b^{-1}\in H\). For \(a,b\in H\), \(ab=a(b^{-1})^{-1}\in H\).
One-step subgroup test (Tag FF-QVQXM)
Let \(G\) be a group and let \(H\subseteq G\) be a nonempty subset. If \(ab^{-1}\in H\) for all \(a,b\in H\), then \(H\) is a subgroup of \(G\).
Choose \(a\in H\); then \(e=aa^{-1}\in H\). For \(b\in H\), \(b^{-1}=eb^{-1}\in H\). For \(a,b\in H\), \(ab=a(b^{-1})^{-1}\in H\).
Intersections, unions, and products of subgroups (Tag E-MSDCC)
Show that the intersection of two subgroups is again a subgroup.
- Hint: one-step subgroup test.
Show that if \(H\cong C_m\) and \(K\cong C_n\) are cyclic subgroups of \(G\), then \(|H\intersect K|\) divides \(\gcd(m,n)\). If \(H\) and \(K\) lie in a common cyclic subgroup of \(G\), show that \(H\intersect K\cong C_d\) where \(d=\gcd(m,n)\).
Show that the intersection of two subgroups with coprime orders is trivial.
Show that the union of two subgroups \(H,K\) is a subgroup iff \(H\subseteq K\) or \(K\subseteq H\), and so is generally not a subgroup.
Show that subgroups with the same prime order are either equal or intersect trivially.
Important for Sylow theory: show (perhaps by example) that if \(S_1, S_2\) are distinct subgroups of order \(p^k\), then it’s possible for their intersection to be trivial or for them to intersect in a subgroup of order \(p^\ell\) for \(1\leq \ell \leq k-1\).
Give a counterexample where \(H,K\leq G\) but \(HK\) is not a subgroup of \(G\).
Solution
<1>1. The intersection of two subgroups is a subgroup.
Let \(H,K\le G\). The identity lies in both subgroups, so \(H\cap K\) is nonempty. If \(x,y\in H\cap K\), then \(x,y\in H\) and \(x,y\in K\). Since both are subgroups, \[ xy^{-1}\in H \qquad\text{and}\qquad xy^{-1}\in K. \] Hence \(xy^{-1}\in H\cap K\). The one-step subgroup test gives \(H\cap K\le G\).
<1>2. If \(H\cong C_m\) and \(K\cong C_n\), then \[ |H\cap K|\mid\gcd(m,n). \] If moreover \(H\) and \(K\) lie in a common cyclic subgroup, then \[ H\cap K\cong C_{\gcd(m,n)}. \]
By <1>1, \(H\cap K\) is a subgroup of both \(H\) and \(K\). Lagrange’s theorem therefore gives \[ |H\cap K|\mid m, \qquad |H\cap K|\mid n, \] so \(|H\cap K|\mid\gcd(m,n)\).
Now suppose \(H,K\le C\) for a cyclic group \(C\). Put \(d=\gcd(m,n)\). A cyclic group has a unique subgroup of each order dividing its order. Let \(D\le C\) be the unique subgroup of order \(d\). Since \(d\mid m\) and \(d\mid n\), one has \(D\le H\) and \(D\le K\), hence \(D\le H\cap K\). The first paragraph gives \(|H\cap K|\le d\), while \(|D|=d\), so \(H\cap K=D\cong C_d\).
<1>3. Subgroups of coprime finite orders intersect trivially.
If \(|H|\) and \(|K|\) are coprime, then by <1>2 the order of \(H\cap K\) divides both \(|H|\) and \(|K|\). Hence \[ |H\cap K|=1, \] so \(H\cap K=\{e\}\).
<1>4. The union \(H\cup K\) is a subgroup if and only if \(H\subseteq K\) or \(K\subseteq H\).
If \(H\subseteq K\), then \(H\cup K=K\); similarly if \(K\subseteq H\).
Conversely, suppose \(H\cup K\) is a subgroup and neither subgroup contains the other. Choose \[ h\in H\setminus K, \qquad k\in K\setminus H. \] Since \(H\cup K\) is assumed to be a subgroup, \(hk\in H\cup K\). If \(hk\in H\), then \[ k=h^{-1}(hk)\in H, \] a contradiction. If \(hk\in K\), then \[ h=(hk)k^{-1}\in K, \] again a contradiction. Thus one subgroup must contain the other.
<1>5. Two subgroups of the same prime order are either equal or intersect trivially.
Let \(|H|=|K|=p\) with \(p\) prime. By Lagrange, \[ |H\cap K|\mid p, \] so \(|H\cap K|\) is either \(1\) or \(p\). In the first case the intersection is trivial. In the second case \(H\cap K\) has the same order as both \(H\) and \(K\), hence \[ H=H\cap K=K. \]
<1>6. Distinct subgroups of order \(p^k\) can have trivial intersection or intersection of order \(p^\ell\) for every \(1\le\ell\le k-1\).
Work in elementary abelian \(p\)-groups, viewed as vector spaces over \(\FF_p\).
For trivial intersection, take \[ G=\FF_p^{2k}, \quad S_1=\spanof\{e_1,\ldots,e_k\}, \quad S_2=\spanof\{e_{k+1},\ldots,e_{2k}\}. \] Then \(|S_1|=|S_2|=p^k\) and \(S_1\cap S_2=0\).
For \(1\le\ell\le k-1\), take \[ G=\FF_p^{2k-\ell}, \] and define \[ S_1=\spanof\{e_1,\ldots,e_k\}, \] \[ S_2=\spanof\{e_1,\ldots,e_\ell,e_{k+1},\ldots,e_{2k-\ell}\}. \] Both have dimension \(k\), hence order \(p^k\), while \[ S_1\cap S_2=\spanof\{e_1,\ldots,e_\ell\} \] has dimension \(\ell\) and order \(p^\ell\).
<1>7. A product \(HK\) of subgroups need not be a subgroup.
Take \(G=S_3\), \[ H=\langle(12)\rangle, \qquad K=\langle(23)\rangle. \] Then \[ HK=\{e,(12),(23),(123)\}, \] so \(|HK|=4\). If \(HK\) were a subgroup of \(S_3\), Lagrange’s theorem would force \(4\mid6\), which is false. Hence \(HK\) is not a subgroup.
Conjugacy
Conjugation is an action of \(G\) on itself and on the set of its subgroups. For the action on elements, the orbit of \(x\) is its conjugacy class and its stabilizer is the centralizer \(C_G(x)\), so \(\size{\operatorname{Cl}(x)}=[G:C_G(x)]\) for finite \(G\). For the action on subgroups, the orbit of \(H\) is the set of conjugate subgroups of \(H\) and its stabilizer is the normalizer \(N_G(H)\), so \(H\) has \([G:N_G(H)]\) conjugates.
Conjugacy class (Tag D-HLDEY)
Let \(G\) be a group and \(h\in G\). The conjugacy class of \(h\) is \[ C(h) \coloneqq \theset{ ghg^{-1} \st g\in G }. \]
Conjugate subgroups (Tag D-ES7MP)
Let \(G\) be a group. Two subgroups \(H, K \leq G\) are conjugate if there exists \(g\in G\) such that \(gHg^{-1} = K\).
Conjugate subgroups have the same cardinality: if \(gHg^{-1} = K\), then \(h\mapsto ghg^{-1}\) is a bijection \(H\to K\).
The conjugates of a proper subgroup of a finite group do not cover the group (Tag E-GYGP2)
Show that if \(H < G\) is a proper subgroup, then \(\Union_{g\in G} gH\inverseof{g} \subset G\) is a proper subset.
Hint: consider the intersection and count. Try Orbit-stabilizer?
Solution
Let \(G\) be finite, \(m=\# H\), and \(n'=[G:H]>1\).
<1>1. \(H\) has \(n=[G:N_G(H)]\le n'\) distinct conjugates.
By orbit-stabilizer for the conjugation action on subgroups, \(n=[G:N_G(H)]\), and \(H\le N_G(H)\) gives \(n\le[G:H]\).
<1>2. \(\# \Union_{g\in G} gH\inverseof{g}\le 1+n(m-1)\).
Each of the \(n\) conjugates has \(m\) elements, one of which is the identity, and the identity is common to all of them.
<1>3. Q.E.D.
By steps <1>1 and <1>2 and Lagrange’s theorem \(n'm=\# G\), \[\# \Union_{g\in G} gH\inverseof{g}\le 1+n'(m-1)=\# G-(n'-1)<\# G .\]
Erratum: the statement needs \(G\) finite. Every matrix in \(\mathrm{GL}_2(\CC)\) is conjugate to an upper-triangular matrix, so the conjugates of the proper subgroup of invertible upper-triangular matrices cover \(\mathrm{GL}_2(\CC)\).
Size of a conjugacy class divides the order of the group (Tag E-VPWK4)
Let \(G\) be a finite group. Show that the size of every conjugacy class in \(G\) divides \(|G|\).
Solution
Let \(x\in G\).
<1>1. Under the conjugation action of \(G\) on itself, the orbit of \(x\) is its conjugacy class and its stabilizer is the centralizer \(C_G(x)\).
The conjugation orbit is \[ G\cdot x=\{gxg^{-1}:g\in G\}=\operatorname{Cl}_G(x). \] Moreover, \[ g\cdot x=x \quad\Longleftrightarrow\quad gxg^{-1}=x \quad\Longleftrightarrow\quad gx=xg, \] so \[ \operatorname{Stab}_G(x)=C_G(x). \]
<1>2. Hence \[ |\operatorname{Cl}_G(x)|=[G:C_G(x)]. \]
This is the orbit-stabilizer theorem applied to the conjugation action, using <1>1.
<1>3. Therefore the size of the conjugacy class divides \(|G|\).
Since \(C_G(x)\le G\) and \(G\) is finite, Lagrange’s theorem gives \[ |G|=|C_G(x)|[G:C_G(x)]. \] By <1>2, the second factor is \(|\operatorname{Cl}_G(x)|\), so \[ |\operatorname{Cl}_G(x)|\mid |G|. \]
The union of conjugates of a proper subgroup is a proper subset (Tag E-ZCJZC)
Show that if \(H < G\) is a proper subgroup, then \(\Union_{g\in G} gH\inverseof{g} \subset G\) is a proper subset.
Hint: consider the intersection and count. Try Orbit-stabilizer?
Solution
Let \(G\) be finite, \(m=\size H\), and \(n'=[G:H]>1\).
<1>1. \(H\) has \(n=[G:N_G(H)]\le n'\) distinct conjugates.
By orbit-stabilizer for the conjugation action on subgroups, \(n=[G:N_G(H)]\), and \(H\le N_G(H)\) gives \(n\le[G:H]\).
<1>2. \(\size \Union_{g\in G} gH\inverseof{g}\le 1+n(m-1)\).
Each of the \(n\) conjugates has \(m\) elements, one of which is the identity, and the identity is common to all of them.
<1>3. Q.E.D.
By steps <1>1 and <1>2 and Lagrange’s theorem \(n'm=\size G\), \[\size \Union_{g\in G} gH\inverseof{g}\le 1+n'(m-1)=\size G-(n'-1)<\size G .\]
Erratum: the statement needs \(G\) finite. Every matrix in \(\mathrm{GL}_2(\CC)\) is conjugate to an upper-triangular matrix, so the conjugates of the proper subgroup of invertible upper-triangular matrices cover \(\mathrm{GL}_2(\CC)\).
Normal subgroups are unions of conjugacy classes (Tag E-BSGSM)
Show that if \(N\trianglelefteq G\) and \(C\) is a conjugacy class in \(G\), then either \(C\subseteq N\) or \(C\cap N=\varnothing\).
Solution
Suppose \(C\cap N\ne\varnothing\), and choose \(h\in C\cap N\). Since \(C\) is the conjugacy class of \(h\), \[ C=\{ghg^{-1}:g\in G\}. \] Normality of \(N\) implies \(ghg^{-1}\in N\) for every \(g\in G\). Hence \(C\subseteq N\).
Therefore every conjugacy class is either contained in \(N\) or disjoint from \(N\).
Size of a conjugacy class equals the index of the centralizer (Tag E-47X7Y)
Prove that the size of a conjugacy class of \(g_i\) is the index of its centralizer, \([G: Z(g_i)] \da [G: C_G(g_i)]\).
Solution
Let \(G\) act on itself by conjugation: \[ h\cdot x=hxh^{-1}. \] The orbit of \(x\) is its conjugacy class \[ \operatorname{Cl}_G(x)=\{hxh^{-1}:h\in G\}, \] and its stabilizer is \[ \operatorname{Stab}_G(x) =\{h:hxh^{-1}=x\} =C_G(x). \]
The orbit-stabilizer bijection is explicitly \[ G/C_G(x)\longrightarrow \operatorname{Cl}_G(x), \qquad hC_G(x)\longmapsto hxh^{-1}. \] Indeed, \[ hxh^{-1}=kxk^{-1} \iff k^{-1}h\in C_G(x) \iff hC_G(x)=kC_G(x). \] Thus \[ \boxed{|\operatorname{Cl}_G(x)|=[G:C_G(x)]}. \] For finite \(G\), this is \(|G|/|C_G(x)|\).
Normal subgroups
A subgroup \(N\le G\) is a normal subgroup if and only if \(gNg^{-1}=N\) for all \(g\in G\), equivalently if and only if \(N\) is a union of conjugacy classes. For \(N\) normal, multiplication of cosets \((aN)(bN)=abN\) is well defined, and \(G/N\) is a group. A nontrivial normal subgroup of a finite $p$-group \(G\) meets the center \(Z(G)\) nontrivially. If \(M,N\normal G\) and \(M\intersect N=1\), then \(mn=nm\) for all \(m\in M\) and \(n\in N\).
Normal subgroup (Tag D-EKE4Q)
Let \(G\) be a group. A subgroup \(N\leq G\) is normal, written \(N\normal G\), if \(gNg^{-1} = N\) for every \(g\in G\).
Let \(G\) be a group and \(N\leq G\). The following are equivalent:
\(N\normal G\).
\(gN = Ng\) for every \(g\in G\).
\(\psi(N) = N\) for every inner automorphism \(\psi \in \Inn(G)\).
\(N\) is the only subgroup of \(G\) conjugate to \(N\).
A subgroup is normal if and only if it is a union of conjugacy classes (Tag PR-PYYRK)
Let \(G\) be a group and \(N\leq G\) a subgroup. Then \(N\) is normal in \(G\) if and only if \(N\) is a union of conjugacy classes of \(G\). In that case \(N=\coprod_{i} C(h_i)\), where \(h_i\) runs over a set of representatives of the conjugacy classes of \(G\) contained in \(N\).
If \(N\normal G\) and \(h\in N\), then \(ghg^{-1}\in N\) for all \(g\in G\), so \(C(h)\subseteq N\) and \(N=\bigcup_{h\in N}C(h)\). Conversely, if \(N\) is a union of conjugacy classes, then for \(h\in N\) and \(g\in G\) we have \(ghg^{-1}\in C(h)\subseteq N\), so \(gNg^{-1}\subseteq N\) for all \(g\); applying this to \(g^{-1}\) gives \(gNg^{-1}=N\). Distinct conjugacy classes are disjoint, so the union is disjoint.
Nontrivial normal subgroups of a $p$-group meet the center (Tag E-EMESP)
Prove that if \(G\) is a \(p\dash\)group, every subgroup \(N\normal G\) intersects the center \(Z(G)\).
Hint: use the class equation.
Solution
Let \(G\) be a finite \(p\)-group and \(1\neq N\normal G\).
<1>1. \(N\) is a disjoint union of \(G\)-conjugacy classes, and the classes of size \(1\) contained in \(N\) are the elements of \(N\cap Z(G)\).
Normality gives \(gng^{-1}\in N\) for \(n\in N\), \(g\in G\), so the \(G\)-conjugacy class of each \(n\in N\) lies in \(N\). The class of \(n\) is \(\{n\}\) exactly when \(n\) commutes with every element of \(G\).
<1>2. Every \(G\)-conjugacy class of size greater than \(1\) has size divisible by \(p\).
The class of \(n\) has size \([G:C_G(n)]\), which divides \(|G|\) and so is a power of \(p\).
<1>3. Q.E.D.
By steps <1>1 and <1>2, \(|N|\equiv|N\cap Z(G)|\pmod p\). Since \(N\neq1\) is a subgroup of a \(p\)-group, \(p\mid|N|\), so \(p\mid|N\cap Z(G)|\). As \(e\in N\cap Z(G)\), this gives \(|N\cap Z(G)|\ge p\), so \(N\cap Z(G)\neq1\).
Trivially intersecting normal subgroups commute (Tag E-LUH54)
Show that if \(H, K \normal G\) and \(H\intersect K = \{e\}\), then \(hk=kh\) for all \(h\in H,k\in K\).
Show that if \(H,K\normal G\) are normal subgroups that intersect trivially, then \([H, K] = 1\) (so \(hk = kh\) for all \(k\) and \(h\)).
Solution
<1>1. For \(h\in H\) and \(k\in K\), the commutator \([h,k]=hkh^{-1}k^{-1}\) lies in \(H\cap K\).
Because \(H\trianglelefteq G\), conjugating \(h^{-1}\) by \(k\) gives \(kh^{-1}k^{-1}\in H\), so \[ [h,k]=h(kh^{-1}k^{-1})\in H. \] Because \(K\trianglelefteq G\), conjugating \(k\) by \(h\) gives \(hkh^{-1}\in K\), so \[ [h,k]=(hkh^{-1})k^{-1}\in K. \] Thus \([h,k]\in H\cap K\).
<1>2. If \(H\cap K=\{e\}\), then \([h,k]=e\) for all \(h\in H\) and \(k\in K\).
By <1>1, every commutator \([h,k]\) lies in \(H\cap K=\{e\}\). Hence \([h,k]=e\).
<1>3. Therefore \(hk=kh\) for all \(h\in H\) and \(k\in K\), equivalently \([H,K]=1\).
The equality \([h,k]=e\) is equivalent to \(hk=kh\). Since this holds for every pair, the subgroup generated by all commutators is trivial.
Normal subgroups of $p$-groups intersect the center (Tag E-P4LG6)
Prove that if \(G\) is a \(p\dash\)group, every subgroup \(N\normal G\) intersects the center \(Z(G)\).
Hint: use the class equation.
Solution
Let \(G\) be a finite \(p\)-group and \(1\neq N\normal G\).
<1>1. \(N\) is a disjoint union of \(G\)-conjugacy classes, and the classes of size \(1\) contained in \(N\) are the elements of \(N\cap Z(G)\).
Normality gives \(gng^{-1}\in N\) for \(n\in N\), \(g\in G\), so the \(G\)-conjugacy class of each \(n\in N\) lies in \(N\). The class of \(n\) is \(\{n\}\) exactly when \(n\) commutes with every element of \(G\).
<1>2. Every \(G\)-conjugacy class of size greater than \(1\) has size divisible by \(p\).
The class of \(n\) has size \([G:C_G(n)]\), which divides \(\size G\) and so is a power of \(p\).
<1>3. Q.E.D.
By steps <1>1 and <1>2, \(\size N\equiv\size(N\cap Z(G))\pmod p\). Since \(N\neq1\) is a subgroup of a \(p\)-group, \(p\mid\size N\), so \(p\mid\size(N\cap Z(G))\). As \(e\in N\cap Z(G)\), this gives \(\size(N\cap Z(G))\ge p\), so \(N\cap Z(G)\neq1\).
Centralizers and centers
For a subset \(S\subseteq G\), the centralizer \(C_G(S)\) consists of the elements fixing every element of \(S\) under conjugation, and the normalizer \(N_G(S)\) of the elements with \(gSg^{-1}=S\); thus \(C_G(S)\le N_G(S)\). The center \(Z(G)=C_G(G)\) is normal in \(G\), and if \(G/Z(G)\) is cyclic then \(G\) is abelian.
Centralizer of an element or subset (Tag D-PX64W)
Let \(G\) be a group and \(h\in G\). The centralizer of \(h\) is \[ Z(h) \coloneqq C_G(h) \coloneqq \theset{ g\in G \st ghg^{-1} = h }. \] For a subset \(H\subseteq G\), the centralizer of \(H\) is \[ Z(H) \coloneqq C_G(H) \coloneqq \theset{g\in G \suchthat ghg^{-1} = h \text{ for all } h\in H} = \Intersect_{h\in H} C_G(h). \]
For the action of \(G\) on itself by conjugation, \(C_G(h)\) is the stabilizer of \(h\), and \(C_G(H)\) is the subgroup of elements that fix every element of \(H\).
Centralizer $Z_G(S)$ of a subset (Tag FD-SSR6N)
Let \(G\) be a group and \(S\subseteq G\) a subset. The centralizer of \(S\) in \(G\) is \[ Z_G(S) \coloneqq \theset{g\in G \st gsg^{-1} = s \text{ for every } s\in S}. \]
Normalizer of a subgroup (Tag D-OZ2RR)
Let \(G\) be a group and \(H\leq G\) a subgroup. The normalizer of \(H\) in \(G\) is \[ N_G(H) \coloneqq \theset{g\in G \suchthat gHg^{-1} = H}. \]
Let \(H\leq G\). Then \(N_G(H)\) is a subgroup of \(G\) in which \(H\) is normal, and every subgroup \(M\leq G\) with \(H\normal M\) is contained in \(N_G(H)\). Hence \[ N_G(H) = \bigcup_{M\in S} M, \qquad S \coloneqq \theset{M\leq G\suchthat H \normal M}. \]
If \(gHg^{-1}=H\) and \(g'Hg'^{-1}=H\), then \((gg')H(gg')^{-1}=H\) and \(g^{-1}Hg=H\), so \(N_G(H)\) is a subgroup; it contains \(H\), and \(H\) is normal in it by definition. If \(H\normal M\), then \(mHm^{-1}=H\) for every \(m\in M\), so \(M\subseteq N_G(H)\). The union of all such \(M\) therefore lies in \(N_G(H)\) and contains \(N_G(H)\in S\).
Conjugation defines a homomorphism \(N_G(H)\to\Aut(H)\), \(g\mapsto(h\mapsto ghg^{-1})\), whose kernel is the centralizer \(C_G(H)\). Thus \(C_G(H)\subseteq N_G(H)\), and elements of \(N_G(H)\) permute the elements of \(H\) by conjugation while elements of \(C_G(H)\) fix them.
Normalizer $N_G(S)$ of a subset (Tag FD-HRCTL)
Let \(G\) be a group and \(S\subseteq G\) a subset, and for \(g\in G\) write \(gSg^{-1}\coloneqq\theset{gsg^{-1} \st s\in S}\). The normalizer of \(S\) in \(G\) is \[ N_G(S) \coloneqq \theset{g\in G \st gSg^{-1} = S}. \]
Center (Tag D-NK7G7)
Let \(G\) be a group. The center of \(G\) is the subgroup \[ Z(G) \coloneqq \theset{ g\in G \st gh = hg \text{ for all } h\in G} = \theset{ g\in G \st C_G(g) = G }, \] where \(C_G(g)\) is the centralizer of \(g\). Its elements are the central elements of \(G\).
$G/Z(G)$ cyclic implies $G$ abelian; $G/N$ abelian iff $[G,G]\le N$ (Tag E-O73XQ)
Show that if \(G/Z(G)\) is cyclic then \(G\) is abelian.
Show that \(G/N\) is abelian iff \([G, G] \leq N\).
Show that normal subgroups are not necessarily contained in \(Z(G)\).
- Hint: consider the order 3 subgroup of \(S_3\).
Solution
If \(G/Z(G)\) is cyclic, then \(G\) is abelian:
Write \(H\da Z(G)\) and \(G/H = \gens{xH}\) as a cyclic quotient.
Fix \(a, b\in G\), then \(aH = x^n H\) and \(bH = x^m H\).
So \(ax^{-n} = h_1, bx^{-m} = h_2\) where the \(h_i\) are now central.
Now write \(ab = (x^n h_1)(x^m h_2) = ba\) by commuting everything.
Cosets
Two left cosets of a subgroup \(H\le G\) are equal or disjoint, so the left cosets partition \(G\). Their number is the index \([G:H]\), and for finite \(G\) the partition gives \(\size G=[G:H]\,\size H\), which is Lagrange’s theorem. For subgroups \(K\le H\le G\) of finite index, \([G:K]=[G:H]\,[H:K]\).
Tower law for subgroups (Tag PR-SF6ZE)
Let \(G\) be a group and \(K\leq H \leq G\) subgroups. Then \[ [G: K] = [G:H]\, [H: K] \] as cardinals; in particular \([G:K]\) is finite if and only if \([G:H]\) and \([H:K]\) are both finite.
Write \(G=\coprod_{i\in I}g_iH\) and \(H=\coprod_{j\in J}h_jK\) as disjoint unions of left cosets, with \(\abs{I}=[G:H]\) and \(\abs{J}=[H:K]\). Then \(G=\bigcup_{i,j}g_ih_jK\). If \(g_ih_jK=g_{i'}h_{j'}K\), then \(g_ih_j\in g_{i'}h_{j'}K\subseteq g_{i'}H\), so \(g_iH=g_{i'}H\) and \(i=i'\); cancelling \(g_i\) gives \(h_jK=h_{j'}K\) and \(j=j'\). Hence the cosets \(g_ih_jK\) are pairwise distinct, and \([G:K]=\abs{I\times J}\).
Larger subgroups have smaller index (Tag PR-YY3JG)
Let \(G\) be a group and \(H\leq K \leq G\) subgroups. Then \[ \abs{G} = [G:1] \geq [G:H] \geq [G:K] \geq [G:G] = 1 \] as cardinals. In particular, for arbitrary subgroups \(H, K\leq G\), applying this to \(H \cap K \leq H\leq G\) and \(H\cap K\leq K\leq G\) gives \([G: H \cap K] \geq [G:H]\) and \([G:H\cap K]\geq[G:K]\).
By the tower law, \([G:H]=[G:K]\,[K:H]\geq[G:K]\) since \([K:H]\geq1\); the outer inequalities are the cases \(H=1\) and \(K=G\).
Cosets are equal or disjoint (Tag PR-VUKHO)
Let \(G\) be a group, \(H\leq G\) a subgroup, and \(x,y\in G\). Then the left cosets \(xH\) and \(yH\) are either equal or disjoint.
Suppose \(g\in xH\cap yH\), say \(g=xh=yh'\) with \(h,h'\in H\). Then \(x=yh'h^{-1}\in yH\), so \(xH\subseteq yHH=yH\); by symmetry \(yH\subseteq xH\).
Equality of left cosets (Tag T-5KGVV)
Let \(G\) be a group, \(H\leq G\) a subgroup, and \(a,b\in G\). Then \[ aH = bH \iff a^{-1} b \in H \iff b^{-1} a\in H. \]
If \(aH=bH\), then \(b=b\cdot e\in aH\), so \(b=ah\) with \(h\in H\) and \(a^{-1}b=h\in H\). If \(a^{-1}b=h\in H\), then \(bH=ahH=aH\). Finally, \(b^{-1}a=(a^{-1}b)^{-1}\), and \(H\) is closed under inverses.
Index of a subgroup (Tag D-VJGH5)
Let \(G\) be a group and \(H\leq G\) a subgroup. The index of \(H\) in \(G\), written \([G:H]\), is the cardinality of the set \(G/H=\theset{gH \st g\in G}\) of left cosets of \(H\).
The map \(gH\mapsto Hg^{-1}\) is a well-defined bijection from left cosets to right cosets, since \(g_1H=g_2H\) if and only if \(g_2^{-1}g_1\in H\) if and only if \(Hg_1^{-1}=Hg_2^{-1}\). So \([G:H]\) is also the number of right cosets of \(H\).
Order of a quotient group (Tag T-WR7CD)
Let \(G\) be a finite group and \(H\normal G\) a normal subgroup. Then \[ [G: H] = \abs{G/H} = \frac{\abs G}{\abs H}. \]
Lagrange's theorem: $|G|/|H|=[G:H]$ for finite $G$ (Tag E-NNTQB)
Show that if \(G\) is finite and \(H\le G\), then \[ [G:H]=\frac{|G|}{|H|}. \]
Solution
The left cosets of \(H\) partition \(G\). For each \(g\in G\), multiplication by \(g\) gives a bijection \[ H\longrightarrow gH,\qquad h\mapsto gh, \] so every coset has \(|H|\) elements. Since there are \([G:H]\) cosets, \[ |G|=[G:H]\,|H|. \] Therefore \[ [G:H]=\frac{|G|}{|H|}. \]
Exercises
Fixed-point-free elements, quotients of $\ZZ^3$ of order $p^2$, groups of order $pq$, and cyclic Sylow $2$-subgroups (Tag E-BAOST)
Show that if \(G\) is a finite group acting transitively on a set \(X\) with at least two elements, then there exists \(g \in G\) which fixes no point of \(X\).
Let \(p\) be prime. For each abelian group \(K\) of order \(p^2\), how many subgroups \(H\leq \ZZ^{\times 3}\) are there with \(\ZZ^3/H \cong K\)?
Let \(\# G = pq\), with \(p,q\) distinct primes. Show that \(G\) has a nontrivial proper normal subgroup, and if \(p\not\equiv 1\mod q\) and \(q\not \equiv 1 \mod p\) then \(G\) is abelian.
Let \(G\) be a finite group and let \(p\) be the smallest prime dividing \(\# G\), and assume \(H\trianglelefteq G\) has order \(p\). Show that \(H \subset Z(G)\).
Let \(G\) be finite of even order and \(P\) a Sylow 2-subgroup. Assume \(P\) is cyclic and generated by \(x\). Show that the sign of the permutation of \(G\) given by left multiplication \(y\mapsto xy\) is \(-1\), and deduce that \(G\) has a nontrivial quotient of order 2.
Solution
<1>1. A finite transitive action on at least two points has a fixed-point-free element.
For \(g\in G\), let \(X^g\) be its fixed-point set. Burnside’s orbit-counting lemma gives \[ 1=|X/G|=\frac1{|G|}\sum_{g\in G}|X^g|, \] so \[ \sum_{g\in G}|X^g|=|G|. \] The identity fixes all of \(X\), and \(|X|\ge2\). If every \(g\in G\) fixed at least one point, then \[ \sum_{g\in G}|X^g| \ge |X|+(|G|-1) \ge |G|+1, \] a contradiction. Hence some \(g\) has \(X^g=\varnothing\).
<1>2. The numbers of subgroups \(H\le\ZZ^3\) with quotient of order \(p^2\) are \[ \begin{array}{c|c} K & \#\{H:\ZZ^3/H\cong K\}\\ \hline \ZZ/p\ZZ\oplus\ZZ/p\ZZ & p^2+p+1,\\ \ZZ/p^2\ZZ & p^2(p^2+p+1). \end{array} \]
There are exactly two abelian groups of order \(p^2\).
For \(K\cong(\ZZ/p\ZZ)^2\), any such \(H\) contains \(p\ZZ^3\), and \(H/p\ZZ^3\) is a one-dimensional subspace of \[ \ZZ^3/p\ZZ^3\cong\FF_p^3. \] Conversely every line in \(\FF_p^3\) gives such an \(H\). The number of lines is \[ \frac{p^3-1}{p-1}=p^2+p+1. \]
For \(K\cong\ZZ/p^2\ZZ\), kernels are counted by surjections \[ \ZZ^3\twoheadrightarrow\ZZ/p^2\ZZ \] modulo postcomposition by automorphisms of the target. A homomorphism is determined by a triple in \((\ZZ/p^2\ZZ)^3\), and it is surjective exactly when the triple is not entirely divisible by \(p\). Thus there are \[ p^6-p^3 \] surjections. Two surjections have the same kernel exactly when they differ by an automorphism of \(\ZZ/p^2\ZZ\), and \[ |\operatorname{Aut}(\ZZ/p^2\ZZ)|=\varphi(p^2)=p(p-1). \] Therefore the number of kernels is \[ \frac{p^6-p^3}{p(p-1)} =p^2\frac{p^3-1}{p-1} =p^2(p^2+p+1). \]
<1>3. Every group of order \(pq\) has a nontrivial proper normal subgroup; under the two stated congruence hypotheses it is abelian.
Interchange \(p\) and \(q\) if necessary so that \(p<q\). Let \(n_q\) be the number of Sylow \(q\)-subgroups. Then \[ n_q\mid p, \qquad n_q\equiv1\pmod q. \] Since \(p<q\), the only possibility is \(n_q=1\). Thus the Sylow \(q\)-subgroup \(Q\) is nontrivial, proper, and normal.
Now assume also \(q\not\equiv1\pmod p\); the other displayed congruence is automatic when \(p<q\). For the number \(n_p\) of Sylow \(p\)-subgroups, \[ n_p\mid q, \qquad n_p\equiv1\pmod p. \] Hence \(n_p=1\) or \(q\), and the congruence hypothesis excludes \(q\), so the Sylow \(p\)-subgroup \(P\) is also normal. Since \(P\cap Q=1\) and \(|P||Q|=|G|\), we have \(G=PQ\). For \(a\in P\) and \(b\in Q\), the commutator \(aba^{-1}b^{-1}\) lies in both \(P\) and \(Q\) because both are normal, hence is trivial. Thus \(P\) and \(Q\) commute elementwise and \(G\cong P\times Q\) is abelian.
<1>4. If \(p\) is the smallest prime divisor of \(|G|\) and \(H\trianglelefteq G\) has order \(p\), then \(H\subseteq Z(G)\).
Since \(|H|=p\), \(H\cong C_p\), so \[ |\operatorname{Aut}(H)|=p-1. \] Normality of \(H\) gives the conjugation homomorphism \[ \theta:G\longrightarrow\operatorname{Aut}(H). \] The order of \(\operatorname{im}\theta\) divides both \(|G|\) and \(p-1\). Every prime divisor of \(p-1\) is smaller than \(p\), while \(p\) is the smallest prime divisor of \(|G|\). Hence \[ \gcd(|G|,p-1)=1, \] so \(\operatorname{im}\theta=1\). Thus every \(g\in G\) centralizes every \(h\in H\), proving \(H\subseteq Z(G)\).
<1>5. If \(G\) has even order and its Sylow \(2\)-subgroup \(P=\langle x\rangle\) is cyclic, then left multiplication by \(x\) has sign \(-1\), and \(G\) has a quotient of order \(2\).
Write \[ |P|=2^r\quad(r\ge1), \qquad |G|=2^r m \] with \(m\) odd. In the left regular action of \(G\) on itself, the permutation \[ \lambda_x:G\to G, \qquad y\longmapsto xy, \] has orbits equal to the left action of \(\langle x\rangle\) on \(G\). Since \(x\) has order \(2^r\), every orbit is a cycle of length \(2^r\). There are \[ |G|/2^r=m \] such cycles. Each \(2^r\)-cycle has sign \[ (-1)^{2^r-1}=-1, \] and \(m\) is odd, so \[ \operatorname{sgn}(\lambda_x)=(-1)^m=-1. \]
The left regular representation \[ \lambda:G\longrightarrow S_G \] is a homomorphism, so \[ \varepsilon:=\operatorname{sgn}\circ\lambda:G\longrightarrow\{\pm1\} \] is a homomorphism. Since \(\varepsilon(x)=-1\), it is surjective. Therefore its kernel is normal of index \(2\), and \[ G/\ker\varepsilon\cong C_2. \]
Fixed-point-free elements, quotients of $\mathbb{Z}^3$, groups of order $pq$, and cyclic Sylow $2$-subgroups (Tag E-BH6Q6)
Show that if \(G\) is a finite group acting transitively on a set \(X\) with at least two elements, then there exists \(g \in G\) which fixes no point of \(X\).
Let \(p\) be prime. For each abelian group \(K\) of order \(p^2\), how many subgroups \(H\leq \ZZ^{\times 3}\) are there with \(\ZZ^3/H \cong K\)?
Let \(\size G = pq\), with \(p,q\) distinct primes. Show that \(G\) has a nontrivial proper normal subgroup, and if \(p\not\equiv 1\mod q\) and \(q\not \equiv 1 \mod p\) then \(G\) is abelian.
Let \(G\) be a finite group and let \(p\) be the smallest prime dividing \(\size G\), and assume \(G\) has a normal subgroup of order \(p\). Show that \(H \subset Z(G)\).
Let \(G\) be finite and \(P\) a Sylow 2-subgroup. Assume \(P\) is cyclic and generated by \(x\). Show that the sign of the permutation of \(G\) corresponding to \(x\mapsto gx\) is 1, and deduce that \(G\) has a nontrivial quotient of order 2.
Solution
<1>1. A finite group acting transitively on a set \(X\) with \(|X|\ge2\) has an element with no fixed point.
For \(g\in G\) let \(X^g\) be its fixed-point set. By Burnside’s orbit-counting lemma, \(1=|X/G|=\frac1{|G|}\sum_{g\in G}|X^g|\), so \(\sum_{g}|X^g|=|G|\). If every \(g\) had \(|X^g|\ge1\), then, since \(|X^e|=|X|\ge2\), \[\sum_{g\in G}|X^g|\ge 2+(|G|-1)=|G|+1,\] a contradiction.
<1>2. The subgroups \(H\le\ZZ^3\) with \(\ZZ^3/H\cong K\) number \(p^2+p+1\) for \(K\cong(\ZZ/p\ZZ)^2\) and \(p^2(p^2+p+1)\) for \(K\cong\ZZ/p^2\ZZ\).
These are the two abelian groups of order \(p^2\). If \(\ZZ^3/H\cong(\ZZ/p\ZZ)^2\), then \(p\ZZ^3\subseteq H\) and \(H/p\ZZ^3\) is a line in \(\ZZ^3/p\ZZ^3\cong\FF_p^3\); conversely each line gives such an \(H\). There are \((p^3-1)/(p-1)=p^2+p+1\) lines.
If \(\ZZ^3/H\cong\ZZ/p^2\ZZ\), then \(H\) is the kernel of a surjection \(\ZZ^3\to\ZZ/p^2\ZZ\), and two surjections have the same kernel exactly when they differ by an automorphism of \(\ZZ/p^2\ZZ\). A homomorphism is a triple in \((\ZZ/p^2\ZZ)^3\), surjective exactly when some entry is a unit, so there are \(p^6-p^3\) surjections. Dividing by \(|\Aut(\ZZ/p^2\ZZ)|=p(p-1)\) gives \(p^2(p^2+p+1)\) kernels.
<1>3. If \(|G|=pq\) with \(p<q\) primes, then the Sylow \(q\)-subgroup \(Q\) is normal; if moreover \(q\not\equiv1\pmod p\), then \(G\cong\ZZ/pq\ZZ\).
\(n_q\mid p\) and \(n_q\equiv1\pmod q\) with \(p<q\) force \(n_q=1\). If \(q\not\equiv1\pmod p\), then \(n_p\mid q\) and \(n_p\equiv1\pmod p\) force \(n_p=1\), so the Sylow \(p\)-subgroup \(P\) is normal too. Then \(P\cap Q=1\), \(G=PQ\), and for \(a\in P\), \(b\in Q\) the commutator \(aba^{-1}b^{-1}\) lies in \(P\cap Q=1\). So \(G\cong P\times Q\cong\ZZ/p\ZZ\times\ZZ/q\ZZ\cong\ZZ/pq\ZZ\).
<1>4. If \(p\) is the smallest prime dividing \(|G|\) and \(H\trianglelefteq G\) has order \(p\), then \(H\subseteq Z(G)\).
Conjugation gives a homomorphism \(\theta\colon G\to\Aut(H)\cong(\ZZ/p\ZZ)^\times\), a group of order \(p-1\). \(|\operatorname{im}\theta|\) divides \(|G|\) and \(p-1\); every prime factor of \(p-1\) is less than \(p\), so \(\gcd(|G|,p-1)=1\) and \(\theta\) is trivial. Thus every element of \(G\) commutes with every element of \(H\).
<1>5. If \(|G|=2^km\) with \(k\ge1\), \(m\) odd, and \(P=\langle x\rangle\) is a cyclic Sylow \(2\)-subgroup, then \(\lambda_x\colon y\mapsto xy\) has sign \(-1\) and \(G\) has a quotient of order \(2\).
The orbits of \(\langle x\rangle\) acting on \(G\) by left multiplication are the right cosets \(Py\), each a cycle of \(\lambda_x\) of length \(2^k\). There are \(m\) of them, each of sign \((-1)^{2^k-1}=-1\), so \(\operatorname{sgn}(\lambda_x)=(-1)^m=-1\). The left regular representation \(\lambda\colon G\to\operatorname{Sym}(G)\) is a homomorphism, so \(\operatorname{sgn}\circ\lambda\colon G\to\{\pm1\}\) is a homomorphism, surjective since it sends \(x\) to \(-1\). Its kernel has index \(2\).
In the last part, the relevant permutation is \(y\mapsto xy\), and its sign is \(-1\), not \(1\); the argument needs \(|G|\) even, since for \(|G|\) odd the Sylow \(2\)-subgroup is trivial and \(G\) has no quotient of order \(2\).