Topics
Recognition of direct products and semidirect products
Product formula: \(\size{HK} = \size H \size K / \size{H\intersect K}\) for subgroups \(H,K\le G\)
Group actions
Orbit-stabilizer
The class equation
Burnside’s formula
Standard actions
Self-action by left translation (the left-regular action)
The homomorphism \(G\to\Sym(G)\), \(g\mapsto \psi_g\) with \(\psi_g(x) \da gx\), is the Cayley representation, a permutation representation of \(G\); \(\Sym(G) \cong S_n\) for \(n\da \size G\)
Self-action by conjugation
Action on the set of subgroups by conjugation
Action on the left cosets \(G/H\) of a fixed subgroup \(H\) by left translation
Transitive subgroups
- The Galois group of an irreducible polynomial of degree \(n\) as a transitive subgroup of \(S_n\)
The structure theorem for finitely generated abelian groups
Invariant factors
Elementary divisors
Simple groups
Automorphisms
Inner automorphisms
Outer automorphisms
Characteristic subgroups
Series of groups
Normal series
Central series
The Jordan–Hölder theorem
- Composition series
Solvable groups
- Derived series
Nilpotent groups
Lower central series
Upper central series
Exercises
Warmup
$HK$ is a subgroup when $H\le N_G(K)$; a counterexample when $K$ is not a subgroup (Tag P-O2H7G)
Let \(H,K\le G\) and suppose \(H\le N_G(K)\). Prove that \(HK\) is a subgroup of \(G\). Give a counterexample if \(K\) is only a subset, not a subgroup.
Solution
<1>1. If \(H\le N_G(K)\), then \(HK\le G\).
The identity belongs to \(HK\). Let \(h_1k_1,h_2k_2\in HK\). Since \(h_2\in N_G(K)\), \[ k_1h_2=h_2(h_2^{-1}k_1h_2) \] with \(h_2^{-1}k_1h_2\in K\). Hence \[ (h_1k_1)(h_2k_2)=h_1h_2\,(h_2^{-1}k_1h_2)k_2\in HK. \] Also, for \(hk\in HK\), \[ (hk)^{-1}=k^{-1}h^{-1}=h^{-1}(hk^{-1}h^{-1})\in HK \] because \(h^{-1}\in N_G(K)\). Therefore \(HK\) is a subgroup.
<1>2. The subgroup hypothesis on \(K\) is necessary.
Take the additive group \(G=\ZZ\), let \(H=2\ZZ\), and let \(K=\{0,1\}\). Since \(G\) is abelian, every element normalizes every subset in the setwise sense, but \[ H+K=2\ZZ\cup(2\ZZ+1)=\ZZ. \] This particular choice still gives a subgroup, so choose instead \(H=\{0\}\) and \(K=\{0,1\}\). Then \(H+K=\{0,1\}\) is not a subgroup of \(\ZZ\) because \(1+1=2\notin\{0,1\}\).
Recognizing direct products (Tag P-P4KA6)
Let \(H,K\trianglelefteq G\) satisfy \[ H\cap K=\{e\},\qquad HK=G. \] Prove that \(G\cong H\times K\). Generalize the criterion to finitely many normal subgroups.
Solution
<1>1. The subgroups \(H\) and \(K\) commute elementwise.
For \(h\in H\) and \(k\in K\), the commutator \[ [h,k]=hkh^{-1}k^{-1} \] lies in \(H\) because \(H\) is normal, and lies in \(K\) because \(K\) is normal. Hence \[ [h,k]\in H\cap K=\{e\}. \] Thus \(hk=kh\).
<1>2. Multiplication gives an isomorphism.
Define \[ \mu:H\times K\to G,\qquad (h,k)\mapsto hk. \] By <1>1 it is a homomorphism. It is surjective because \(HK=G\). If \(hk=e\), then \(h=k^{-1}\in H\cap K\), so \(h=k=e\). Thus \(\mu\) is injective.
<1>3. Finite-family version.
Let \(H_1,\dots,H_r\trianglelefteq G\). If \[ G=H_1\cdots H_r \] and for every \(i\), \[ H_i\cap \prod_{j\ne i}H_j=\{e\}, \] then the subgroups commute pairwise and the multiplication map \[ H_1\times\cdots\times H_r\longrightarrow G \] is an isomorphism.
Equivalently, the internal direct-product condition is precisely that the multiplication map be surjective with trivial kernel. Pairwise trivial intersections alone are not sufficient for three or more factors.
Group action, orbit, stabilizer, and fixed points (Tag P-WC3HY)
State definitions of the following:
Group action
Orbit
Stabilizer
Fixed points
Solution
Let \(G\) be a group acting on a set \(X\).
A group action is a map \[ G\times X\to X,\qquad (g,x)\mapsto g\cdot x, \] such that \[ e\cdot x=x, \qquad (gh)\cdot x=g\cdot(h\cdot x). \]
The orbit of \(x\in X\) is \[ Gx=\{g\cdot x:g\in G\}. \]
The stabilizer of \(x\) is \[ G_x=\{g\in G:g\cdot x=x\}. \]
For \(g\in G\), its fixed-point set is \[ X^g=\{x\in X:g\cdot x=x\}. \] The fixed points of the whole action are \[ X^G=\{x\in X:g\cdot x=x\text{ for every }g\in G\}. \]
Orbit-stabilizer theorem (Tag P-5YUW6)
- State the orbit-stabilizer theorem
Solution
Let a group \(G\) act on a set \(X\), and let \(x\in X\). Write \[ G_x=\{g\in G:g\cdot x=x\} \] for the stabilizer and \(G\cdot x\) for the orbit.
<1>1. The map \[ \Phi:G/G_x\longrightarrow G\cdot x, \qquad gG_x\longmapsto g\cdot x \] is a bijection.
It is well-defined: if \(gG_x=hG_x\), then \(h^{-1}g\in G_x\), so \[ g\cdot x=h(h^{-1}g)\cdot x=h\cdot x. \] It is surjective by the definition of the orbit. If \(g\cdot x=h\cdot x\), then \[ h^{-1}g\cdot x=x, \] so \(h^{-1}g\in G_x\) and therefore \(gG_x=hG_x\). Hence it is injective.
<1>2. Therefore \[ |G\cdot x|=[G:G_x] \] whenever these cardinalities are finite.
By <1>1, the orbit is in bijection with the left coset space \(G/G_x\).
<1>3. If \(G\) is finite, then \[ |G|=|G\cdot x|\,|G_x|. \]
By Lagrange’s theorem, \[ |G|=[G:G_x]|G_x|. \] Substitute <1>2. This is the orbit–stabilizer theorem.
The class equation (Tag P-JZM4U)
State the class equation and derive it from orbit-stabilizer.
Solution
Let a finite group \(G\) act on itself by conjugation, \[ g\cdot x=gxg^{-1}. \] The orbit of \(x\) is its conjugacy class, and its stabilizer is its centralizer \[ C_G(x)=\{g\in G:gx=xg\}. \] By orbit-stabilizer, \[ |\operatorname{Cl}_G(x)|=[G:C_G(x)]. \]
The conjugacy classes partition \(G\). The classes of size \(1\) are exactly the central elements, because \[ |\operatorname{Cl}_G(x)|=1 \iff C_G(x)=G \iff x\in Z(G). \] Choosing one representative \(x_i\) from each noncentral conjugacy class gives the class equation \[ |G|=|Z(G)|+\sum_i [G:C_G(x_i)]. \]
The center of a $p$-group is nontrivial (Tag P-3PD4W)
Prove that the centre of a group of order \(p^r\) (\(p\) prime) is not trivial.
Solution
Let \(|G|=p^r\). Under conjugation, the class equation is \[ |G|=|Z(G)|+\sum_i [G:C_G(x_i)], \] where the \(x_i\) represent the noncentral conjugacy classes.
<1>1. Every noncentral conjugacy-class size is divisible by \(p\).
For noncentral \(x_i\), the centralizer \(C_G(x_i)\) is a proper subgroup of the \(p\)-group \(G\). Hence \[ [G:C_G(x_i)]>1 \] and, being a divisor of \(|G|=p^r\), it is a positive power of \(p\). Therefore it is divisible by \(p\).
<1>2. The order \(|Z(G)|\) is divisible by \(p\).
Reducing the class equation modulo \(p\) and using <1>1 gives \[ 0\equiv |G|\equiv |Z(G)|\pmod p. \] Thus \(p\mid |Z(G)|\).
<1>3. Therefore \(Z(G)\) is nontrivial.
Since \(p\mid |Z(G)|\), one has \(|Z(G)|\ge p>1\). Hence \(Z(G)\ne\{e\}\).
Classification of abelian groups of a given composite order (Tag P-2TASE)
Important: Pick your favorite composite number \(m = \prod p_i^{e_i}\) and classify all abelian groups of that order.
- Write their invariant factor decompositions and their elementary divisor decompositions. Come up with an algorithm for converting back and forth between these.
Solution
Take \[ m=72=2^3\cdot3^2. \] By the classification theorem for finite abelian groups, the \(2\)-primary part is determined by a partition of \(3\), and the \(3\)-primary part by a partition of \(2\).
<1>1. The possible primary components are \[ \begin{aligned} G_{(2)}&\cong C_8,\quad C_4\oplus C_2,\quad C_2\oplus C_2\oplus C_2,\\ G_{(3)}&\cong C_9,\quad C_3\oplus C_3. \end{aligned} \]
The partitions of \(3\) are \[ 3,\qquad2+1,\qquad1+1+1, \] which give the three \(2\)-primary groups. The partitions of \(2\) are \[ 2,\qquad1+1, \] which give the two \(3\)-primary groups.
<1>2. Hence there are exactly six abelian groups of order \(72\).
The primary decomposition is unique, so each choice of a \(2\)-primary type and a \(3\)-primary type gives one isomorphism class, and every class occurs this way. Thus there are \[ 3\cdot2=6 \] classes.
<1>3. Their elementary-divisor and invariant-factor forms are as follows: \[ \begin{array}{c|c} \text{elementary divisors}&\text{invariant factors}\\ \hline C_8\oplus C_9&C_{72}\\ C_8\oplus C_3\oplus C_3&C_3\oplus C_{24}\\ C_4\oplus C_2\oplus C_9&C_2\oplus C_{36}\\ C_4\oplus C_2\oplus C_3\oplus C_3&C_6\oplus C_{12}\\ C_2\oplus C_2\oplus C_2\oplus C_9&C_2\oplus C_2\oplus C_{18}\\ C_2\oplus C_2\oplus C_2\oplus C_3\oplus C_3&C_2\oplus C_6\oplus C_6 \end{array} \]
For relatively prime integers \(a,b\), one has \[ C_a\oplus C_b\cong C_{ab}. \] To obtain invariant factors, align the prime-power cyclic factors in nondecreasing order, pad the shorter lists on the left by \(1\)’s, and multiply columnwise. For example, \[ (C_4,C_2)=(C_2,C_4), \qquad (C_9)=(C_1,C_9), \] so \[ (C_2\oplus C_4)\oplus C_9 \cong C_{2\cdot1}\oplus C_{4\cdot9} =C_2\oplus C_{36}. \] The other rows follow identically. In every invariant-factor row, each factor divides the next.
<1>4. The conversion algorithm from elementary divisors to invariant factors is columnwise multiplication of aligned prime powers.
For each prime \(p\), write the \(p\)-primary cyclic factors as \[ p^{a_{p,1}}\mid p^{a_{p,2}}\mid\cdots\mid p^{a_{p,r_p}}. \] Let \(r=\max_p r_p\) and pad each list on the left by exponent \(0\). For each column \(j\), define \[ d_j=\prod_p p^{a_{p,j}}. \] Then \[ d_1\mid d_2\mid\cdots\mid d_r, \] and the direct sum of the \(C_{d_j}\) is the invariant-factor decomposition.
<1>5. The reverse algorithm factors each invariant factor into its prime-power parts.
Given \[ C_{d_1}\oplus\cdots\oplus C_{d_r}, \qquad d_1\mid\cdots\mid d_r, \] factor each \[ d_j=\prod_p p^{v_p(d_j)}. \] Using the Chinese remainder theorem, \[ C_{d_j}\cong\bigoplus_{p\mid d_j}C_{p^{v_p(d_j)}}. \] Collecting these prime-power cyclic summands gives the elementary-divisor decomposition. The two algorithms are inverse because they merely regroup the same primary cyclic factors.
A finite group is not the union of conjugates of a proper subgroup; $S_n$ has a derangement (Tag P-6XW74)
Prove that if \(G\) is a finite group and \(H<G\) is a proper subgroup, then \(G\) cannot be written as a union of conjugates of \(H\).
Use this to prove that if \(G=\Sym(X)\) is the group of permutations of a finite set \(X\) with \(|X|=n>1\), then there exists \(g\in G\) with no fixed point in \(X\).
Solution
<1>1. Let \(G\) act transitively on the coset space \(X=G/H\) by left multiplication. Then an element \(g\in G\) fixes some coset if and only if it lies in a conjugate of \(H\).
A coset \(aH\) is fixed by \(g\) exactly when \[ gaH=aH, \] which is equivalent to \[ a^{-1}ga\in H. \] Equivalently, \[ g\in aHa^{-1}. \] Thus the elements lying in at least one conjugate of \(H\) are exactly the elements having at least one fixed point in the coset action.
<1>2. Every finite transitive action on a set with more than one point has a fixed-point-free element.
Let a finite group \(G\) act transitively on a finite set \(X\) with \(|X|>1\). For \(g\in G\), write \(X^g\) for the 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 contributes \(|X|>1\). If every nonidentity element fixed at least one point, then \[ \sum_{g\in G}|X^g| \ge |X|+(|G|-1)>|G|, \] a contradiction. Hence some element has no fixed points.
<1>3. A finite group is not the union of the conjugates of a proper subgroup.
If \(H<G\), then the coset space \(G/H\) has more than one point and the action is transitive. By <1>2, some \(g\in G\) fixes no coset. By <1>1, this \(g\) belongs to no conjugate of \(H\). Therefore \[ \bigcup_{a\in G}aHa^{-1}\ne G. \]
<1>4. For every finite set \(X\) with \(|X|=n>1\), the symmetric group \(\Sym(X)\) contains a derangement.
Fix \(x\in X\) and let \[ H=\operatorname{Stab}_{\Sym(X)}(x). \] This is a proper subgroup. Since \(\Sym(X)\) acts transitively on \(X\), the conjugates of \(H\) are exactly the point stabilizers \[ \operatorname{Stab}_{\Sym(X)}(y),\qquad y\in X. \] By <1>3, choose \[ g\notin\bigcup_{y\in X}\operatorname{Stab}(y). \] Then \(g\) fixes no point of \(X\), so \(g\) is a derangement.
Composition series, solvability, and nilpotence (Tag P-HYQCG)
Define what a composition series is, and state what it means for a group to be simple, solvable, or nilpotent.
- How are the derived and lower/upper central series defined? What type(s) of the groups above does each series correspond to?
Solution
A nontrivial group is simple if its only normal subgroups are \(1\) and itself.
A composition series of a finite group \(G\) is a subnormal series \[G=G_0\triangleright G_1\triangleright\cdots\triangleright G_n=1\] in which each factor \(G_i/G_{i+1}\) is simple; equivalently, each \(G_{i+1}\) is a maximal proper normal subgroup of \(G_i\). By the Jordan–Hölder theorem, the multiset of composition factors, up to isomorphism, does not depend on the composition series.
The derived series is \(G^{(0)}=G\), \(G^{(i+1)}=[G^{(i)},G^{(i)}]\), and \(G\) is solvable if \(G^{(n)}=1\) for some \(n\). Each \(G^{(i)}/G^{(i+1)}\) is abelian; conversely, if \(G=H_0\trianglerighteq H_1\trianglerighteq\cdots\trianglerighteq H_r=1\) has abelian factors, induction gives \(G^{(i)}\le H_i\), so \(G^{(r)}=1\). Thus \(G\) is solvable exactly when it has a finite subnormal series with abelian factors.
The lower central series is \(\gamma_1(G)=G\), \(\gamma_{i+1}(G)=[\gamma_i(G),G]\), and the upper central series is \(Z_0(G)=1\), \(Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G))\). The group \(G\) is nilpotent if \(\gamma_{c+1}(G)=1\) for some \(c\); equivalently, \(Z_c(G)=G\) for some \(c\), and the least such \(c\) is the same for both series, the nilpotency class.
Thus composition series record the simple factors, the derived series detects solvability, and the lower and upper central series detect nilpotence. Every nilpotent group is solvable, since \(G^{(i)}\le\gamma_{i+1}(G)\); the converse fails for \(S_3\), which is solvable with trivial center.
Group actions
Orbits, stabilizers, and kernels of the standard $G$-actions (Tag P-N2TM2)
For each standard action below, identify the orbits, stabilizers, global fixed points, kernel, and image in the relevant symmetric group:
- left translation of \(G\) on itself;
- conjugation of \(G\) on itself;
- conjugation of \(G\) on its set of subgroups;
- left translation of \(G\) on \(G/H\) for a fixed subgroup \(H\le G\).
Solution
<1>1. Left translation on \(G\).
The action is transitive, so there is one orbit, namely \(G\). The stabilizer of every \(x\in G\) is trivial because \[ gx=x\iff g=e. \] Hence the kernel is trivial and the image is the left regular copy of \(G\) in \(\Sym(G)\). If \(G\ne1\), there are no global fixed points.
<1>2. Conjugation on \(G\).
The orbit of \(x\) is its conjugacy class \[ \operatorname{Cl}_G(x). \] Its stabilizer is the centralizer \[ C_G(x). \] The global fixed points are exactly \(Z(G)\). The kernel consists of elements conjugating every \(x\) trivially, hence is also \(Z(G)\). Therefore the image is \[ \Inn(G)\cong G/Z(G). \]
<1>3. Conjugation on the set of subgroups.
The orbit of \(H\) is the conjugacy class of the subgroup, \[ \{gHg^{-1}:g\in G\}. \] The stabilizer is its normalizer \[ N_G(H). \] The global fixed points are precisely the normal subgroups of \(G\). The kernel is \[ \bigcap_{H\le G}N_G(H), \] the subgroup of elements normalizing every subgroup of \(G\). Thus the image is the quotient of \(G\) by this kernel.
<1>4. Left translation on \(G/H\).
The action is transitive. The stabilizer of the coset \(xH\) is \[ xHx^{-1}. \] The kernel is the core of \(H\), \[ \operatorname{core}_G(H) =\bigcap_{x\in G}xHx^{-1}, \] the largest normal subgroup of \(G\) contained in \(H\). Hence the image is \[ G/\operatorname{core}_G(H). \] If \(H<G\), there is no global fixed coset; if \(H=G\), the unique coset is fixed.
Stabilizer subgroups of two points in a $G$-set are conjugate (Tag P-YMVNK)
Let \(X\) be a \(G\)-set with group action \(G \times X \to X\). (1) Prove that if \(x_1, x_2 \in X\) belong to the same \(G\)-orbit (\(x_2 = g \cdot x_1\) for some \(g \in G\)), then their stabilizer subgroups \(\operatorname{Stab}_G(x_1)\) and \(\operatorname{Stab}_G(x_2)\) are conjugate in \(G\): \[\operatorname{Stab}_G(g \cdot x_1) = g \operatorname{Stab}_G(x_1) g^{-1}.\] (2) Note that if \(x_1, x_2\) lie in different orbits, their stabilizers need not be conjugate.
Solution
For \(h\in G\), \[ \begin{aligned} h\in\operatorname{Stab}_G(gx) &\iff h(gx)=gx\\ &\iff (g^{-1}hg)x=x\\ &\iff g^{-1}hg\in\operatorname{Stab}_G(x)\\ &\iff h\in g\operatorname{Stab}_G(x)g^{-1}. \end{aligned} \] Hence \[ \operatorname{Stab}_G(gx)=g\operatorname{Stab}_G(x)g^{-1}. \] Thus points in the same orbit have conjugate stabilizers.
For different orbits this can fail. For example, let \(S_3\) act on \[ \{1,2,3\}\sqcup\{*\} \] in the usual way on the first orbit and trivially on \(*\). Then \[ \operatorname{Stab}(1)\cong C_2, \qquad \operatorname{Stab}(*)=S_3, \] so the stabilizers are not conjugate.
A subgroup of index equal to the smallest prime dividing $|G|$ is normal (Tag P-BQVDL)
Let \(G\) be a finite group, and let \(H \le G\) be a subgroup of index \([G : H] = p\), where \(p\) is the smallest prime dividing \(|G|\). Prove that \(H\) is a normal subgroup of \(G\) (\(H \trianglelefteq G\)).
Solution
Let \(G\) act on the \(p\) left cosets of \(H\). This gives \[ \rho:G\to S_p, \] and let \(K=\ker\rho\). Then \(K\trianglelefteq G\), \(K\le H\), and \[ [G:K]=|\operatorname{im}\rho|\mid p!. \] Since \([G:H]=p\), \[ [G:K]=p[H:K], \] so \([H:K]\mid (p-1)!\).
On the other hand, \([H:K]\mid |G|\). If \([H:K]>1\), choose a prime \(q\mid[H:K]\). Then \(q\mid|G|\), so by minimality of \(p\) we have \(q\ge p\). But \(q\mid(p-1)!\), so \(q<p\), a contradiction. Hence \([H:K]=1\), so \(H=K\trianglelefteq G\).
Infinite simple groups have no finite-index subgroups (Tag P-KOQPF)
Show that if \(G\) is an infinite simple group, then \(G\) cannot have a proper subgroup of finite index.
Solution
Suppose \(H<G\) has finite index \(n\). The action of \(G\) on the left cosets \(G/H\) gives a homomorphism \[ \rho:G\to S_n. \] Its kernel is the core \[ \ker\rho=\bigcap_{g\in G}gHg^{-1}, \] which is a normal subgroup of \(G\) contained in \(H\).
If \(G\) is simple, then \(\ker\rho\) is either \(1\) or \(G\). Since \(H\) is proper, \(\ker\rho\ne G\), so \(\ker\rho=1\). Hence \(\rho\) is injective and embeds \(G\) into the finite group \(S_n\), forcing \(G\) to be finite. This contradicts the hypothesis that \(G\) is infinite.
Therefore an infinite simple group has no proper subgroup of finite index.
Subgroups of order 5 in $S_5$ are transitive (Tag P-Q44SQ)
Show that every subgroup of order \(5\) in \(S_5\) acts transitively on \(\{1,2,3,4,5\}\).
Solution
Let \(H\le S_5\) have order \(5\). Then \(H\) is cyclic, say \[ H=\langle\sigma\rangle \] with \(\sigma\ne e\). Since \(\sigma\) has order \(5\), its cycle decomposition in \(S_5\) must consist of a single \(5\)-cycle. Therefore the orbit of any point under \(\langle\sigma\rangle\) has size \(5\) and is the whole set.
Hence \(H\) is transitive.
Automorphisms
Euler totient of primes and composites (Tag P-BXBHJ)
- How do you compute the totient \(\phi(p)\) for \(p\) prime? Or \(\phi(n)\) for \(n\) composite?
Solution
<1>1. If \(p\) is prime, then \[ \varphi(p)=p-1. \]
Every nonzero residue modulo \(p\) is relatively prime to \(p\), so the invertible residue classes are exactly \(1,2,\ldots,p-1\).
<1>2. For a prime power \(p^a\) with \(a\ge1\), \[ \varphi(p^a)=p^a-p^{a-1}=p^{a-1}(p-1). \]
Among the \(p^a\) residue classes modulo \(p^a\), exactly the \(p^{a-1}\) multiples of \(p\) fail to be relatively prime to \(p^a\).
<1>3. Euler’s totient is multiplicative on coprime integers.
If \(\gcd(m,n)=1\), the Chinese remainder theorem gives \[ \ZZ/mn\ZZ\cong \ZZ/m\ZZ\times\ZZ/n\ZZ. \] Taking unit groups gives \[ (\ZZ/mn\ZZ)^\times\cong(\ZZ/m\ZZ)^\times\times(\ZZ/n\ZZ)^\times, \] so \[ \varphi(mn)=\varphi(m)\varphi(n). \]
<1>4. Therefore, if \[ n=\prod_{i=1}^r p_i^{a_i} \] is the prime factorization, then \[ \varphi(n)=\prod_{i=1}^r p_i^{a_i-1}(p_i-1) =n\prod_{p\mid n}\left(1-\frac1p\right). \]
Apply <1>2 to each prime-power factor and then <1>3.
Order of $\GL_n(\FF_q)$ (Tag P-BJDIE)
Show that the order of the group \(\mathrm{GL}_n(\mathbb{F}_q)\) of invertible \(n\times n\) matrices over the field \(\mathbb{F}_q\) of \(q\) elements is \((q^n-1)(q^n-q)\cdots(q^n-q^{n-1})\).
Solution
An invertible \(n\times n\) matrix over \(\FF_q\) is the same thing as an ordered basis of the vector space \(\FF_q^n\), namely its ordered list of columns.
<1>1. The first column can be any nonzero vector, giving \(q^n-1\) choices.
There are \(q^n\) vectors in \(\FF_q^n\), and only the zero vector is forbidden.
<1>2. After choosing \(k\) linearly independent columns, the next column has \(q^n-q^k\) choices.
The span of \(k\) independent vectors is a \(k\)-dimensional subspace, hence has \(q^k\) elements. The next column must lie outside that span.
<1>3. Therefore \[ |\GL_n(\FF_q)|=(q^n-1)(q^n-q)(q^n-q^2)\cdots(q^n-q^{n-1}). \]
Choose the columns successively and multiply the numbers of choices from <1>1 and <1>2.
$\Aut(\ZZ/p)$, $\Aut((\ZZ/p)^n)$, and $\Aut(\ZZ/n)$ (Tag P-TRZ3O)
Identify \[ \operatorname{Aut}(\ZZ/p\ZZ), \qquad \operatorname{Aut}((\ZZ/p\ZZ)^n), \qquad \operatorname{Aut}(\ZZ/n\ZZ). \]
Solution
For the additive cyclic group \(C_p=\ZZ/p\ZZ\), every automorphism is determined by the image of \(1\), which may be any nonzero residue class. Thus \[ \operatorname{Aut}(\ZZ/p\ZZ)\cong(\ZZ/p\ZZ)^\times. \] Since \(p\) is prime, this is a cyclic group of order \(p-1\).
The group \[ (\ZZ/p\ZZ)^n \] is the additive group of the \(n\)-dimensional vector space \(\FF_p^n\). Its group automorphisms are exactly the invertible \(\FF_p\)-linear maps, so \[ \operatorname{Aut}((\ZZ/p\ZZ)^n)\cong GL_n(\FF_p). \]
Finally, for the additive cyclic group \(\ZZ/n\ZZ\), an endomorphism is determined by \[ 1\longmapsto a\pmod n. \] It is invertible exactly when \(a\) is a unit modulo \(n\). Therefore \[ \operatorname{Aut}(\ZZ/n\ZZ)\cong(\ZZ/n\ZZ)^\times. \]
Number of elements of order $4$ in $\Aut(\ZZ/20)$ (Tag P-SUU6D)
How many elements in the automorphism group \(\operatorname{Aut}(\mathbb{Z}/20\mathbb{Z})\) have order 4?
Solution
Since \(\mathbb Z/20\mathbb Z\) is cyclic, \[ \operatorname{Aut}(\mathbb Z/20\mathbb Z)\cong(\mathbb Z/20\mathbb Z)^\times. \] By the Chinese remainder theorem, \[ (\mathbb Z/20\mathbb Z)^\times \cong(\mathbb Z/4\mathbb Z)^\times\times(\mathbb Z/5\mathbb Z)^\times \cong C_2\times C_4. \] An element \((a,b)\in C_2\times C_4\) has order \(4\) exactly when \(b\) has order \(4\). There are two choices for \(a\) and two elements of order \(4\) in \(C_4\), hence \[ 2\cdot2=4 \] such elements.
Explicitly, they correspond to \[ 3,7,13,17\pmod{20}. \] Therefore the answer is \[ \boxed{4}. \]
Non-isomorphic groups can have isomorphic automorphism groups (Tag P-BIFDZ)
- Find two groups \(G\not\cong H\) where \(\Aut G\cong \Aut H\).
Solution
Take \[ G=C_3, \qquad H=C_4. \] Then \(G\not\cong H\) because they have different orders. On the other hand, \[ \Aut(C_n)\cong (\ZZ/n\ZZ)^\times. \] Therefore \[ \Aut(C_3)\cong (\ZZ/3\ZZ)^\times\cong C_2 \] and \[ \Aut(C_4)\cong (\ZZ/4\ZZ)^\times\cong C_2. \] Hence \[ \Aut(G)\cong\Aut(H) \] even though \(G\not\cong H\).
Isomorphic subgroups need not have isomorphic quotients (Tag P-3OUVI)
Let \(H, K \leq G\) be subgroups with \(H\cong K\). Is it true that \(G/H \cong G/K\)?
Hint: consider a group with distinct subgroups of order 2 whose quotients have order 4.
Solution
No.
<1>1. Let \[ G=C_4\oplus C_2, \] and write elements additively. Define \[ H=\langle(2,0)\rangle, \qquad K=\langle(0,1)\rangle. \] Then \(H\cong K\cong C_2\).
Both displayed generators have order \(2\). Since \(G\) is abelian, both subgroups are normal, so both quotients are defined.
<1>2. The quotient \(G/H\) is isomorphic to \(C_2\oplus C_2\).
Quotienting the \(C_4\) factor by its subgroup of order \(2\) gives \(C_2\), while the second factor survives unchanged: \[ G/H\cong (C_4/\langle2\rangle)\oplus C_2\cong C_2\oplus C_2. \]
<1>3. The quotient \(G/K\) is isomorphic to \(C_4\).
The subgroup \(K\) is exactly the second direct factor, so \[ G/K\cong C_4. \]
<1>4. Hence \(G/H\not\cong G/K\).
The group \(C_4\) contains an element of order \(4\), whereas every nonzero element of \(C_2\oplus C_2\) has order \(2\). Thus the quotients are not isomorphic even though \(H\cong K\).
Inner automorphisms preserve conjugacy of subgroups (Tag P-HPSTB)
Show that any automorphism (and in particular, every inner automorphism) sends conjugate subgroups to conjugate subgroups.
Solution
Suppose \(H_2=gH_1g^{-1}\) and let \(\varphi\in\operatorname{Aut}(G)\). Then \[ \varphi(H_2) =\varphi(gH_1g^{-1}) =\varphi(g)\,\varphi(H_1)\,\varphi(g)^{-1}. \] Thus \(\varphi(H_1)\) and \(\varphi(H_2)\) are conjugate subgroups of \(G\). In particular, this holds for every inner automorphism.
$\Aut(S_n)=\Inn(S_n)$ for $n\neq 6$ (Tag P-WBK5B)
Show that for \(n\ne6\), \[ \operatorname{Aut}(S_n)=\operatorname{Inn}(S_n). \]
Solution
For \(n=1,2\) the statement is immediate, so assume \(n\ge3\).
<1>1. Every automorphism sends transpositions to transpositions when \(n\ne6\).
An involution in \(S_n\) has cycle type \(2^k1^{n-2k}\). Its conjugacy class has size \[ \frac{n!}{2^k k!(n-2k)!}. \] A transposition has class size \(\binom n2\). Automorphisms preserve element order and conjugacy-class size, so the image of a transposition must be an involution satisfying \[ \frac{n!}{2^k k!(n-2k)!}=\binom n2. \] For \(k\ge2\) this is equivalent to \[ \frac{(n-2)!}{2^{k-1}k!(n-2k)!}=1. \] For \(k=2\) there is no integer solution. For \(k=3\) the unique solution is \(n=6\). For \(k\ge4\), already at the smallest possible \(n=2k\) the left side is \[ \frac{(2k-2)!}{2^{k-1}k!}>1, \] and it increases with \(n\). Thus, when \(n\ne6\), necessarily \(k=1\).
<1>2. An automorphism preserving transpositions is inner.
Let \[ s_i=(i\ i+1),\qquad 1\le i<n. \] These generate \(S_n\). Put \(t_i=\phi(s_i)\). Each \(t_i\) is a transposition. Moreover, \[ \operatorname{ord}(t_it_j)= \operatorname{ord}(s_is_j) = \begin{cases} 3,&|i-j|=1,\\ 2,&|i-j|>1. \end{cases} \] Two distinct transpositions have product of order \(3\) exactly when their supports meet in one point, and they commute exactly when their supports are disjoint.
Write \(t_1=(a_1a_2)\). Since \(t_2\) meets \(t_1\) in exactly one point, after interchanging \(a_1,a_2\) if necessary write \(t_2=(a_2a_3)\). Inductively, \(t_i\) must meet \(t_{i-1}\) but be disjoint from \(t_1,\ldots,t_{i-2}\), so \[ t_i=(a_i a_{i+1}) \] for distinct \(a_1,\ldots,a_n\).
Let \(\pi\in S_n\) be defined by \(\pi(i)=a_i\). Then \[ \pi s_i\pi^{-1}=t_i=\phi(s_i) \] for every \(i\). Since the \(s_i\) generate \(S_n\), \[ \phi(g)=\pi g\pi^{-1} \qquad(g\in S_n). \] Hence \(\phi\) is inner.
Therefore \[ \operatorname{Aut}(S_n)=\operatorname{Inn}(S_n) \] for all \(n\ne6\).
Series of groups
Pairs $(n,p)$ for which $\SL_n(\FF_p)$ is solvable (Tag P-KPEPM)
Let \(p\) be prime and \(n\ge1\). Determine all pairs \((n,p)\) for which \[ \SL_n(\FF_p) \] is solvable.
Solution
The solvable cases are \[ n=1\quad\text{for every prime }p, \] and \[ (n,p)=(2,2),(2,3). \]
<1>1. If \(n=1\), then \(\SL_1(\FF_p)=1\), so it is solvable.
<1>2. The groups \(\SL_2(\FF_2)\) and \(\SL_2(\FF_3)\) are solvable.
We have \[ \SL_2(\FF_2)\cong S_3, \] which is solvable.
For \(p=3\), the center of \(\SL_2(\FF_3)\) is \(\{\pm I\}\) and \[ \PSL_2(\FF_3)\cong A_4. \] Thus \(\SL_2(\FF_3)\) is an extension of the solvable group \(A_4\) by the abelian group \(C_2\), hence is solvable.
<1>3. All other cases with \(n\ge2\) are nonsolvable.
The classical simplicity theorem for projective special linear groups says that \[ \PSL_n(\FF_p) \] is nonabelian simple for \(n\ge2\), except for \[ (n,p)=(2,2),(2,3). \] Since \(\PSL_n(\FF_p)\) is a quotient of \(\SL_n(\FF_p)\), solvability of \(\SL_n(\FF_p)\) would force solvability of \(\PSL_n(\FF_p)\). A nonabelian simple group is not solvable, so no other pair occurs.
Groups of order $pq$ need not be nilpotent (Tag P-NLMMW)
If \(|G| = pq\) (distinct primes), is \(G\) necessarily nilpotent?
Solution
No. For example, \(S_3\) has order \(6=2\cdot3\) but is not nilpotent: its center is trivial, so its upper central series cannot reach \(S_3\).
More generally, for a finite group, nilpotence is equivalent to normality of every Sylow subgroup. If \(|G|=pq\) with primes \(p<q\), the Sylow \(q\)-subgroup is always normal. A nonabelian group of order \(pq\) exists exactly when \(p\mid(q-1)\); it is a semidirect product \[ C_q\rtimes C_p \] with nontrivial action, and its Sylow \(p\)-subgroups are not normal. Hence such a group is not nilpotent. The case \(p=2,q=3\) is \(S_3\).
Solvable groups have a nontrivial normal subgroup (Tag P-RPSTO)
Show that if \(G\) is solvable, then \(G\) contains a nontrivial normal subroup.
- What does this mean on the Galois theory side?
Hint: consider the derived series.
Solution
Taken literally, every nontrivial group \(G\) contains a nontrivial normal subgroup, namely \(G\) itself. The solvability hypothesis becomes meaningful if one asks for a proper nontrivial normal subgroup.
If \(G\) is nonabelian and solvable, then \[ 1\ne [G,G]\ne G. \] Indeed, \([G,G]\ne1\) because \(G\) is nonabelian, while \([G,G]\ne G\) because otherwise the derived series would be stationary at \(G\) and could not reach \(1\). Since \([G,G]\) is characteristic, it is normal. Thus every nonabelian solvable group has a proper nontrivial normal subgroup.
For abelian solvable groups this stronger assertion can fail: \(C_p\) is simple. More generally, a nontrivial abelian group has a proper nontrivial subgroup unless it has prime order.
For a finite Galois extension \(L/F\) with solvable Galois group \(G\), a normal subgroup \(H\trianglelefteq G\) corresponds to an intermediate field \(L^H\) that is Galois over \(F\). A proper nontrivial normal subgroup gives a proper nontrivial Galois intermediate field. The derived series gives the usual tower whose successive Galois groups are abelian.
Qual problems
A finite group is abelian if conjugacy class representatives pairwise commute (Tag P-P3RNI)
Let \(G\) be a finite group with \(n\) distinct conjugacy classes, and let \(g_1, g_2, \dots, g_n\) be a set of representatives for the conjugacy classes of \(G\).
Prove that if \(g_i g_j = g_j g_i\) for all \(1 \le i, j \le n\), then \(G\) is abelian.
Solution
Goal: Prove that \(G\) is abelian by considering the subgroup \(H = \langle g_1, \dots, g_n \rangle\) and applying the conjugate cover theorem for finite groups.
<1>1. Subgroup \(H = \langle g_1, \dots, g_n \rangle\) generated by representatives:
<2>1. Define \(H = \langle g_1, \dots, g_n \rangle \le G\) to be the subgroup of \(G\) generated by the \(n\) conjugacy class representatives. <2>2. By hypothesis, \(g_i g_j = g_j g_i\) for all \(1 \le i, j \le n\), so all generators of \(H\) pairwise commute. <2>3. Since any group generated by pairwise commuting elements is abelian, \(H\) is an abelian subgroup of \(G\).
<1>2. \(G\) is the union of conjugates of \(H\):
<2>1. Let \(x \in G\) be an arbitrary element. <2>2. Since \(g_1, \dots, g_n\) represent all conjugacy classes of \(G\), the conjugacy classes partition \(G\): \[G = \bigsqcup_{i=1}^n \operatorname{Cl}(g_i).\] <2>3. Thus \(x \in \operatorname{Cl}(g_k)\) for some \(k \in \{1, \dots, n\}\). <2>4. By definition of conjugacy class, there exists an element \(g \in G\) such that \(x = g g_k g^{-1}\). <2>5. Since \(g_k \in H\), \(x = g g_k g^{-1} \in g H g^{-1}\). <2>6. Therefore: \[G \subseteq \bigcup_{g \in G} g H g^{-1} \subseteq G \implies G = \bigcup_{g \in G} g H g^{-1}.\]
<1>3. Lemma: No finite group is a union of conjugates of a proper subgroup.
<2>1. Let \(K < G\) be a proper subgroup with \([G : K] = m > 1\). <2>2. The number of distinct conjugate subgroups of \(K\) in \(G\) is \([G : N_G(K)] \le [G : K] = m\). <2>3. Each conjugate \(g K g^{-1}\) contains \(|K|\) elements and shares the identity element \(e\) with all other conjugates. <2>4. The union of all conjugates contains at most \[\left| \bigcup_{g \in G} g K g^{-1} \right| \le 1 + [G : N_G(K)] (|K| - 1) \le 1 + [G : K] (|K| - 1) = 1 + |G| - [G : K].\] <2>5. Since \([G : K] \ge 2\), \(1 + |G| - [G : K] \le |G| - 1 < |G|\). <2>6. Thus \(\bigcup_{g \in G} g K g^{-1} \subsetneq G\).
<1>4. \(H = G\) and \(G\) is abelian:
<2>1. From <1>2, \(G = \bigcup_{g \in G} g H g^{-1}\). <2>2. By the contrapositive of <1>3, \(H\) cannot be a proper subgroup of \(G\). <2>3. Thus \(H = G\). <2>4. From <1>1, \(H\) is abelian, so \(G\) is abelian.
<1>5. Conclusion:
$G$ is abelian.$p$-groups have nontrivial center, groups of order $p^2$ and $5^2\cdot 7^2$ are abelian, and the isomorphism classes of groups of order $5^2\cdot 7^2$ (Tag P-TR6YB)
Use the Class Equation to prove that any finite \(p\)-group (a group of order \(p^k\) for a prime \(p\) and integer \(k \ge 1\)) has a non-trivial center.
Prove that any group of order \(p^2\) (where \(p\) is prime) is abelian.
Prove that any group of order \(5^2 \cdot 7^2 = 1225\) is abelian.
Write down exactly one representative from each isomorphism class of groups of order \(5^2 \cdot 7^2\).
Solution
Goal: Prove non-triviality of the center for \(p\)-groups via the class equation in (a), deduce abelianness of order \(p^2\) in (b), establish direct product decomposition of order \(5^2 \cdot 7^2\) via Sylow theory in (c), and enumerate isomorphism classes in (d).
<1>1. Part (a): Finite \(p\)-groups have non-trivial centers.
<2>1. Let \(G\) be a group of order \(|G| = p^k\) with \(k \ge 1\). <2>2. \(G\) acts on itself by conjugation (\(g \cdot x = g x g^{-1}\)). <2>3. The orbit of \(x \in G\) is the conjugacy class \(\operatorname{Cl}(x)\), and the stabilizer of \(x\) is the centralizer \(C_G(x)\). <2>4. By the Orbit-Stabilizer Theorem, \(|\operatorname{Cl}(x)| = [G : C_G(x)]\). <2>5. The conjugacy class has size 1 if and only if \(g x g^{-1} = x\) for all \(g \in G\), which is equivalent to \(x \in Z(G)\). <2>6. Partitioning \(G\) into disjoint conjugacy classes yields the Class Equation: \[|G| = |Z(G)| + \sum_{i=1}^m [G : C_G(x_i)],\] where \(x_1, \dots, x_m\) are representatives of the conjugacy classes of size strictly greater than 1. <2>7. For each \(i \in \{1, \dots, m\}\), \([G : C_G(x_i)] > 1\) divides \(|G| = p^k\), so \(p\) divides \([G : C_G(x_i)]\). <2>8. Rearranging the Class Equation: \[|Z(G)| = |G| - \sum_{i=1}^m [G : C_G(x_i)].\] <2>9. Since \(p \mid |G|\) and \(p \mid [G : C_G(x_i)]\) for each \(i\), \(p\) divides \(|Z(G)|\). <2>10. Since \(e \in Z(G)\), \(|Z(G)| \ge 1\). Since \(p \mid |Z(G)|\), \(|Z(G)| \ge p > 1\). <2>11. Thus \(Z(G)\) is non-trivial.
<1>2. Part (b): Every group of order \(p^2\) is abelian.
<2>1. Let \(|G| = p^2\). By Part (a), \(|Z(G)| \in \{p, p^2\}\). <2>2. If \(|Z(G)| = p^2\), then \(Z(G) = G\), so \(G\) is abelian. <2>3. If \(|Z(G)| = p\), then the quotient group \(G/Z(G)\) has order \(|G/Z(G)| = p^2 / p = p\). <2>4. Every group of prime order is cyclic, so \(G/Z(G)\) is cyclic, generated by some coset \(x Z(G)\). <2>5. Lemma: If \(G/Z(G)\) is cyclic, then \(G\) is abelian: - Every element \(g \in G\) can be written as \(g = x^a z_1\) for some \(a \in \mathbb{Z}\) and \(z_1 \in Z(G)\). - For any two elements \(g = x^a z_1\) and \(h = x^b z_2\): \[g h = (x^a z_1)(x^b z_2) = x^a x^b z_1 z_2 = x^{a+b} z_2 z_1 = x^b x^a z_2 z_1 = (x^b z_2)(x^a z_1) = h g.\] - Thus \(G\) is abelian. <2>6. But if \(G\) is abelian, \(Z(G) = G\), so \(|Z(G)| = p^2 \ne p\), a contradiction. <2>7. Therefore \(|Z(G)| = p^2\), which proves \(G\) is abelian.
<1>3. Part (c): Every group of order \(5^2 \cdot 7^2\) is abelian.
<2>1. Let \(|G| = 5^2 \cdot 7^2 = 1225\). <2>2. By Sylow’s Theorems: - Number of Sylow 5-subgroups: \(n_5 \equiv 1 \pmod 5\) and \(n_5 \mid 7^2 = 49\). Divisors of 49 are \(\{1, 7, 49\}\). Among these, only \(1 \equiv 1 \pmod 5\), so \(n_5 = 1\). - Number of Sylow 7-subgroups: \(n_7 \equiv 1 \pmod 7\) and \(n_7 \mid 5^2 = 25\). Divisors of 25 are \(\{1, 5, 25\}\). Among these, only \(1 \equiv 1 \pmod 7\), so \(n_7 = 1\). <2>3. Let \(P\) be the unique Sylow 5-subgroup and \(Q\) be the unique Sylow 7-subgroup. <2>4. Both \(P\) and \(Q\) are normal in \(G\) (\(P \trianglelefteq G, Q \trianglelefteq G\)). <2>5. Since \(\gcd(|P|, |Q|) = \gcd(25, 49) = 1\), \(P \cap Q = \{e\}\). <2>6. Since \(|P Q| = |P||Q| = 25 \cdot 49 = 1225 = |G|\), \(G\) is the internal direct product: \[G \cong P \times Q.\] <2>7. By Part (b), since \(|P| = 5^2\) and \(|Q| = 7^2\), both \(P\) and \(Q\) are abelian. <2>8. The direct product of abelian groups is abelian, so \(G\) is abelian.
<1>4. Part (d): Isomorphism classes of groups of order \(5^2 \cdot 7^2\).
<2>1. By Part (c), every group of order \(5^2 \cdot 7^2\) is abelian of the form \(P \times Q\), where \(P\) has order 25 and \(Q\) has order 49. <2>2. By the Fundamental Theorem of Finite Abelian Groups: - Abelian groups of order \(5^2 = 25\): \(\mathbb{Z}/25\mathbb{Z}\) and \(\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}\). - Abelian groups of order \(7^2 = 49\): \(\mathbb{Z}/49\mathbb{Z}\) and \(\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z}\). <2>3. Combining these yields exactly \(2 \times 2 = 4\) non-isomorphic abelian groups: 1. \(\mathbb{Z}/25\mathbb{Z} \times \mathbb{Z}/49\mathbb{Z} \cong \mathbb{Z}/1225\mathbb{Z}\) 2. \((\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}) \times \mathbb{Z}/49\mathbb{Z}\) 3. \(\mathbb{Z}/25\mathbb{Z} \times (\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z})\) 4. \((\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}) \times (\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z})\)
<1>5. Conclusion:
$p$-groups have $|Z(G)| \ge p$, order $p^2$ groups are abelian, order $5^2 \cdot 7^2$ groups decompose into $P \times Q$ (hence abelian), with exactly 4 isomorphism classes.Orbit-stabilizer bijection $G/G_x\simeq G\cdot x$, and nontriviality of $Z(G)$ for finite $p$-groups (Tag P-FKAJJ)
Let \(G\) be a finite group acting on a set \(X\). For \(x\in X\), let \(G_x\) be the stabilizer of \(x\) and \(G\cdot x\) be the orbit of \(x\).
Prove that there is a bijection between the left cosets \(G/G_x\) and \(G\cdot x\).
Prove that the center of every finite \(p\dash\)group \(G\) is nontrivial by considering that action of \(G\) on \(X=G\) by conjugation.
Solution
For (a), define \[ \Phi:G/G_x\longrightarrow G\cdot x, \qquad \Phi(gG_x)=g\cdot x. \] If \(gG_x=hG_x\), then \(h^{-1}g\in G_x\), so \((h^{-1}g)\cdot x=x\) and therefore \(g\cdot x=h\cdot x\). Thus \(\Phi\) is well defined. It is surjective by the definition of the orbit. If \(g\cdot x=h\cdot x\), then \(h^{-1}g\in G_x\), so \(gG_x=hG_x\); hence \(\Phi\) is injective. Therefore it is a bijection, and in particular \[ |G\cdot x|=[G:G_x]. \]
For (b), let the finite \(p\)-group \(G\) act on itself by conjugation. The fixed points are exactly the elements of \(Z(G)\). For any \(x\in G\), the stabilizer is the centralizer \(C_G(x)\), so by part (a) the conjugacy class of \(x\) has size \[ [G:C_G(x)], \] a power of \(p\). Hence every non-singleton conjugacy class has size divisible by \(p\). The class equation therefore gives \[ |G|=|Z(G)|+\sum_i |\mathcal C_i|, \] where every term in the sum is divisible by \(p\). Since \(p\mid |G|\), it follows that \(p\mid |Z(G)|\). The identity lies in \(Z(G)\), so \(|Z(G)|\ge p\), and in particular \(Z(G)\) is nontrivial.
Stabilizers in an orbit are conjugate, conjugates of a proper subgroup do not cover $G$, and a transitive action on two or more points has a fixed-point-free element (Tag P-EI5VA)
Suppose the group \(G\) acts on a set \(X\). Show that the stabilizers of elements in the same orbit are conjugate: for any \(x \in X\) and \(g \in G\), \(G_{g \cdot x} = g G_x g^{-1}\).
Let \(G\) be a finite group and let \(H < G\) be a proper subgroup. Show that the union of all conjugates of \(H\) is strictly smaller than \(G\): \[ \bigcup_{g \in G} g H g^{-1} \subsetneq G. \]
Suppose a finite group \(G\) acts transitively on a set \(S\) with \(|S| \ge 2\). Show that there exists an element \(g \in G\) having no fixed points in \(S\) (i.e. \(g \cdot s \ne s\) for all \(s \in S\)).
Solution
Goal: Prove conjugacy of stabilizers in (a), prove that conjugates of a proper subgroup do not cover \(G\) in (b), and deduce the existence of a fixed-point-free element (derangement) for transitive actions in (c).
<1>1. Part (a): Conjugacy of stabilizers in an orbit.
<2>1. Let \(x \in X\) and \(g \in G\), and let \(y = g \cdot x \in X\). <2>2. An element \(h \in G\) stabilizes \(y\) if and only if \(h \cdot y = y\). <2>3. Substituting \(y = g \cdot x\): \[h \cdot (g \cdot x) = g \cdot x \iff (g^{-1} h g) \cdot x = x \iff g^{-1} h g \in G_x.\] <2>4. Multiplying on the left by \(g\) and on the right by \(g^{-1}\): \[g^{-1} h g \in G_x \iff h \in g G_x g^{-1}.\] <2>5. Therefore \(G_y = G_{g \cdot x} = g G_x g^{-1}\).
<1>2. Part (b): Conjugates of a proper subgroup do not cover \(G\).
<2>1. Let \(N_G(H) = \{g \in G : g H g^{-1} = H\}\) be the normalizer of \(H\) in \(G\). <2>2. The distinct conjugates of \(H\) in \(G\) are parameterized by the cosets of \(N_G(H)\), so there are precisely \(k = [G : N_G(H)]\) distinct conjugates \(H_1, H_2, \dots, H_k\). <2>3. Since \(H \le N_G(H) \le G\), by the tower law of indices \([G : H] = [G : N_G(H)] [N_G(H) : H] \ge [G : N_G(H)] = k\). <2>4. Every conjugate \(g H g^{-1}\) contains the identity element \(e\). <2>5. Each of the \(k\) distinct conjugates contains \(|H| - 1\) non-identity elements. <2>6. Bound the size of the union of all conjugates: \[\left| \bigcup_{g \in G} g H g^{-1} \right| = \left| \{e\} \cup \bigcup_{i=1}^k (H_i \setminus \{e\}) \right| \le 1 + \sum_{i=1}^k (|H_i| - 1) = 1 + k (|H| - 1).\] <2>7. Since \(k \le [G : H]\), we have \[\left| \bigcup_{g \in G} g H g^{-1} \right| \le 1 + [G : H](|H| - 1) = 1 + [G : H]|H| - [G : H] = |G| + 1 - [G : H].\] <2>8. Since \(H\) is a proper subgroup (\(H \ne G\)), the index \([G : H] \ge 2\). <2>9. Thus: \[\left| \bigcup_{g \in G} g H g^{-1} \right| \le |G| + 1 - [G : H] \le |G| - 1 < |G|.\] <2>10. Therefore \(\bigcup_{g \in G} g H g^{-1} \subsetneq G\).
<1>3. Part (c): Transitive actions on \(|S| \ge 2\) have a fixed-point-free element.
<2>1. Method 1 (via Part (b)): <3>1. Choose an arbitrary point \(s_0 \in S\), and let \(H = G_{s_0} \le G\) be its stabilizer. <3>2. By the Orbit-Stabilizer Theorem, since the action is transitive, \([G : H] = |G \cdot s_0| = |S| \ge 2\). <3>3. Thus \(H\) is a proper subgroup of \(G\). <3>4. By Part (a), every stabilizer is of the form \(G_{g \cdot s_0} = g H g^{-1}\). <3>5. An element \(g \in G\) fixes at least one point \(s \in S\) if and only if \(g \in G_s = h H h^{-1}\) for some \(h \in G\). <3>6. Thus the set of elements in \(G\) that fix at least one point of \(S\) is precisely \(\bigcup_{h \in G} h H h^{-1}\). <3>7. By Part (b), \(\bigcup_{h \in G} h H h^{-1} \subsetneq G\). <3>8. Therefore there exists an element \(g \in G \setminus \bigcup_{h \in G} h H h^{-1}\), which fixes no points of \(S\). <2>2. Method 2 (via Burnside’s Lemma): <3>1. Let \(X^g = \{s \in S : g \cdot s = s\}\) denote the fixed point set of \(g \in G\). <3>2. By Burnside’s Lemma, the number of orbits is the average number of fixed points: \[1 = \frac{1}{|G|} \sum_{g \in G} |X^g| \implies |G| = \sum_{g \in G} |X^g| = |X^e| + \sum_{g \ne e} |X^g|.\] <3>3. Since the identity fixes all of \(S\), \(|X^e| = |S| \ge 2\). <3>4. If every \(g \ne e\) had at least one fixed point (\(|X^g| \ge 1\)), then \[\sum_{g \in G} |X^g| = |X^e| + \sum_{g \ne e} |X^g| \ge 2 + (|G| - 1) = |G| + 1,\] a contradiction. <3>5. Thus there exists at least one \(g \in G\) with \(|X^g| = 0\).
<1>4. Conclusion:
Stabilizers in an orbit are conjugate, conjugates of a proper subgroup cannot cover a finite group, and transitive actions on sets of size $\ge 2$ always contain derangements.Faithful transitive actions: trivial core of a stabilizer, and abelian transitive subgroups of $S_n$ (Tag P-DXHST)
Suppose the group \(G\) acts on the non-empty set \(A\). Assume this action is faithful (\(\ker(G \to \operatorname{Sym}(A)) = \{1\}\)) and transitive (for all \(a, b \in A\), there exists \(g \in G\) such that \(g \cdot a = b\)).
For \(a \in A\), let \(G_a = \operatorname{Stab}_G(a)\) denote the stabilizer of \(a\). Prove that: \[ \bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{1\}. \]
Suppose that \(G\) is abelian. Prove that \(|G| = |A|\), and deduce that every abelian transitive subgroup of the symmetric group \(S_n\) has order exactly \(n\).
Solution
Goal: Prove that the kernel of a transitive action is the normal core \(\bigcap_{g \in G} g G_a g^{-1} = \{1\}\), and show that transitive faithful abelian actions are regular (\(G_a = \{1\}\) and \(|G| = |A|\)).
<1>1. Part (a): Proof that the Normal Core \(\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{1\}\): Proof: <2>1. The action of \(G\) on \(A\) is given by a group homomorphism \(\rho: G \to \operatorname{Sym}(A)\), where \(\rho(g)(x) = g \cdot x\). <2>2. The kernel of the action is defined as: \[\ker(\rho) = \{ g \in G \mid g \cdot x = x \text{ for all } x \in A \} = \bigcap_{x \in A} G_x.\] <2>3. Since the action is transitive, every element \(x \in A\) can be written as \(x = \sigma \cdot a\) for some \(\sigma \in G\). <2>4. The stabilizer of \(x = \sigma \cdot a\) is the conjugate subgroup: \[G_x = G_{\sigma \cdot a} = \sigma G_a \sigma^{-1}.\] <2>5. Therefore, the intersection of all stabilizers over all points in \(A\) is: \[\bigcap_{x \in A} G_x = \bigcap_{\sigma \in G} \sigma G_a \sigma^{-1}.\] <2>6. Since the action is faithful by hypothesis, \(\ker(\rho) = \{1\}\). <2>7. Thus: \[\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \ker(\rho) = \{1\}.\]
<1>2. Part (b): Faithful Transitive Actions of Abelian Groups are Regular: Proof: <2>1. Now assume that \(G\) is abelian. <2>2. Since \(G\) is abelian, every subgroup of \(G\) is normal. <2>3. In particular, the stabilizer subgroup \(G_a\) is normal in \(G\): \[\sigma G_a \sigma^{-1} = G_a \quad \text{for all } \sigma \in G.\] <2>4. Substituting this into the intersection formula from Part (a): \[\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \bigcap_{\sigma \in G} G_a = G_a.\] <2>5. By Part (a), this intersection is trivial: \[G_a = \{1\}.\] <2>6. By the Orbit-Stabilizer Theorem, since the action is transitive, the single orbit is the entire set \(A = G \cdot a\): \[|A| = [G : G_a] = \frac{|G|}{|G_a|} = \frac{|G|}{1} = |G|.\] <2>7. Thus \(|G| = |A|\) (the action is free and transitive, i.e., simply transitive / regular).
<1>3. Application to Abelian Transitive Subgroups of \(S_n\): Proof: <2>1. Let \(G \le S_n\) be an abelian transitive subgroup acting on \(A = \{1, 2, \dots, n\}\). <2>2. Since \(G\) is a subgroup of \(S_n = \operatorname{Sym}(A)\), the permutation action of \(G\) on \(A\) is faithful by definition (\(\rho: G \hookrightarrow S_n\) is an inclusion). <2>3. Since \(G\) is abelian and transitive, by Step 2 we have: \[|G| = |A| = n.\]
<1>4. Conclusion: (a) \(\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \ker(\rho) = \{1\}\); (b) \(G_a = \{1\} \implies |G| = |A| = n\). Q.E.D.
Using the Sylow theorems
Groups of order $105$: normal Sylow $5$- and $7$-subgroups, a cyclic subgroup of order $35$, and cyclicity when the Sylow $3$-subgroup is normal (Tag P-VI6QM)
Let \(G\) be a group of order \(105 = 3 \cdot 5 \cdot 7\), and let \(P \in \operatorname{Syl}_3(G)\), \(Q \in \operatorname{Syl}_5(G)\), and \(R \in \operatorname{Syl}_7(G)\) be Sylow 3-, 5-, and 7-subgroups of \(G\), respectively.
Prove that at least one of \(Q\) and \(R\) is normal in \(G\).
Prove that \(G\) has a cyclic subgroup of order 35.
Prove that both \(Q\) and \(R\) are normal in \(G\).
Prove that if \(P\) is normal in \(G\), then \(G\) is cyclic.
Solution
Goal: Prove normality of Sylow subgroups, existence of a cyclic subgroup of order 35, and deduce that \(G\) is cyclic when all Sylow subgroups are normal.
<1>1. Part (a): At least one of \(Q\) or \(R\) is normal in \(G\).
<2>1. By Sylow’s Theorems, the number \(n_5\) of Sylow 5-subgroups satisfies \(n_5 \equiv 1 \pmod 5\) and \(n_5 \mid (3 \cdot 7) = 21\). Thus \(n_5 \in \{1, 21\}\). <2>2. The number \(n_7\) of Sylow 7-subgroups satisfies \(n_7 \equiv 1 \pmod 7\) and \(n_7 \mid (3 \cdot 5) = 15\). Thus \(n_7 \in \{1, 15\}\). <2>3. Suppose for contradiction that neither \(Q\) nor \(R\) is normal in \(G\), so \(n_5 = 21\) and \(n_7 = 15\). <2>4. Count elements: - Each Sylow 5-subgroup has prime order 5, so any two distinct Sylow 5-subgroups intersect only in the identity \(\{e\}\). Together they contain \(21 \cdot (5 - 1) = 84\) elements of order 5. - Each Sylow 7-subgroup has prime order 7, so any two distinct Sylow 7-subgroups intersect only in \(\{e\}\). Together they contain \(15 \cdot (7 - 1) = 90\) elements of order 7. - Since 5 and 7 are distinct primes, no element can have both order 5 and order 7. <2>5. The identity, the elements of order 5, and the elements of order 7 give at least \[1 + 84 + 90 = 175 \text{ distinct elements in } G.\] <2>6. This contradicts \(|G| = 105\). <2>7. Thus \(n_5 = 1\) or \(n_7 = 1\), so at least one of \(Q\) or \(R\) is normal in \(G\).
<1>2. Part (b): \(G\) has a cyclic subgroup of order 35.
<2>1. By Part (a), \(Q \trianglelefteq G\) or \(R \trianglelefteq G\). <2>2. If one of two subgroups is normal, their set product is a subgroup. Thus \(H = Q R \le G\). <2>3. By Lagrange’s Theorem, \(|Q \cap R|\) divides \(\gcd(|Q|, |R|) = \gcd(5, 7) = 1\), so \(Q \cap R = \{e\}\). <2>4. The order of \(H\) is \[|H| = \frac{|Q| |R|}{|Q \cap R|} = \frac{5 \cdot 7}{1} = 35.\] <2>5. \(H\) is cyclic: - In \(H\), the number of Sylow 7-subgroups \(n_7(H)\) satisfies \(n_7(H) \equiv 1 \pmod 7\) and \(n_7(H) \mid 5 \implies n_7(H) = 1\), so the Sylow 7-subgroup \(R \trianglelefteq H\). - The number of Sylow 5-subgroups \(n_5(H)\) satisfies \(n_5(H) \equiv 1 \pmod 5\) and \(n_5(H) \mid 7 \implies n_5(H) = 1\), so the Sylow 5-subgroup \(Q \trianglelefteq H\). - Since \(Q, R \trianglelefteq H\) and \(Q \cap R = \{e\}\), \(H \cong Q \times R \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z} \cong \mathbb{Z}/35\mathbb{Z}\). <2>6. Thus \(H\) is a cyclic subgroup of order 35 in \(G\).
<1>3. Part (c): Both \(Q\) and \(R\) are normal in \(G\).
<2>1. Let \(H \le G\) be the cyclic subgroup of order 35 from Part (b). <2>2. The index of \(H\) in \(G\) is \([G : H] = \frac{105}{35} = 3\). <2>3. Since 3 is the smallest prime dividing \(|G| = 105\), any subgroup of index 3 is normal in \(G\). Thus \(H \trianglelefteq G\). <2>4. In the cyclic group \(H \cong \mathbb{Z}/35\mathbb{Z}\), \(Q\) is the unique subgroup of order 5, and \(R\) is the unique subgroup of order 7. <2>5. Because \(Q\) and \(R\) are uniquely determined by their orders in \(H\), they are characteristic subgroups of \(H\): for every automorphism \(\sigma \in \operatorname{Aut}(H)\), \(\sigma(Q) = Q\) and \(\sigma(R) = R\). <2>6. Since \(H \trianglelefteq G\), conjugation by any \(g \in G\) restricts to an automorphism of \(H\). <2>7. Thus \(g Q g^{-1} = Q\) and \(g R g^{-1} = R\) for all \(g \in G\), proving that both \(Q \trianglelefteq G\) and \(R \trianglelefteq G\).
<1>4. Part (d): If \(P \trianglelefteq G\), then \(G\) is cyclic.
<2>1. Assume \(P \trianglelefteq G\). By Part (c), \(Q \trianglelefteq G\) and \(R \trianglelefteq G\). <2>2. All three Sylow subgroups \(P, Q, R\) are normal in \(G\). <2>3. Since \(|P| = 3\), \(|Q| = 5\), \(|R| = 7\) are pairwise coprime: - \(P \cap (Q R) = \{e\}\) because \(|P| = 3\) and \(|Q R| = 35\) with \(\gcd(3, 35) = 1\). - \(Q \cap R = \{e\}\) because \(\gcd(5, 7) = 1\). <2>4. The product \(P Q R\) is a subgroup of \(G\) with order \(|P Q R| = |P| |Q| |R| = 3 \cdot 5 \cdot 7 = 105 = |G|\), so \(G = P Q R\). <2>5. Therefore \(G\) is isomorphic to the internal direct product: \[G \cong P \times Q \times R \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z}.\] <2>6. By the Chinese Remainder Theorem, \(\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z} \cong \mathbb{Z}/105\mathbb{Z}\), which is cyclic.
<1>5. Conclusion:
$Q$ and $R$ are normal in $G$, $G$ contains a cyclic subgroup of order 35, and normality of $P$ implies $G \cong \mathbb{Z}/105\mathbb{Z}$.