Original public practice · AG-1 · October 2026

Abstract algebra I: groups and symmetry

Connect group actions, structure, representations and geometric symmetry in complete arguments. This form contains 10 multipart questions and 100 practice points.

This is undergraduate abstract algebra. It is separate from school Algebra I and II. These original problems sample the published material; they are not an official university paper, a calibrated grade prediction or an examination of every course objective.

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State the ambient group, ring or field and check theorem hypotheses. You may cite results taught in the linked lessons unless a part asks for their proof. Supply exact reasoning and the requested calculations; a theorem name or numerical match alone is not a proof. Alternative valid methods can earn the same points.

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1. Actions and structure

40 practice points.

Question 1: A faithful action that loses information on blocks

10 points

Let S₄ act on the six unordered two-element subsets of {1,2,3,4}. Put N={1,(12)(34),(13)(24),(14)(23)}. Permutations act right to left. Suggested time: 24 minutes, not learner-calibrated.

(a) · 3 points

Prove the action is transitive and faithful. List the stabilizer of {1,2} and check its order using orbit–stabilizer.

(b) · 3 points

Prove N is normal and find its orbits on the six pairs. Determine the image and kernel of the induced action on those orbits.

(c) · 4 points

For a finite transitive H-action on X and a normal subgroup M, prove the M-orbits are blocks of equal size and H acts transitively on their set. A block has every translate either identical or disjoint. Must the block-action kernel equal M? Give a checked counterexample if not.

Review question 1: worked argument and scoring criteria

(a)

Any bijection between two pairs extends to a permutation of the four letters, proving transitivity. The stabilizer is {1,(12),(34),(12)(34)}: it permutes the pair and its complement independently. Its order is 24/6=4. If g fixes every pair, it fixes {i,j}∩{i,k}={i} for distinct i,j,k. Thus it fixes every letter and is identity, proving faithfulness.

1 point: Proves transitivity by extending a pair bijection.
1 point: Lists the complete stabilizer and checks its order.
1 point: Proves the original action has trivial kernel.

(b)

The three nonidentity elements square to 1, and the product of two distinct ones is the third. Conjugation permutes the double transpositions, so N is normal. Its orbits are {{1,2},{3,4}}, {{1,3},{2,4}}, {{1,4},{2,3}}. Normality gives g(Nx)=N(gx), making the induced action well-defined. On these blocks (12) is a transposition and (123) a three-cycle; they generate S₃. The kernel therefore has order 24/6=4. N is contained in it, so it is exactly N.

1 point: Checks subgroup, normality and all three orbits.
1 point: Justifies the induced action and its full S₃ image.
1 point: Identifies the kernel by inclusion and order.

(c)

If y=hx, normality yields h(Mx)=My. Thus translates of an orbit are other orbits, which are identical or disjoint; this is the block property. Transitivity on X gives transitivity on the orbit set, and h gives a bijection between corresponding orbits, proving equal sizes. Every m∈M preserves each orbit, so M is contained in the block kernel. Equality need not hold: let S₃ act on three letters and M=A₃. The subgroup M is proper and transitive, so there is one block and its action has kernel S₃, strictly larger than M. The six-pair example instead has kernel exactly N, even though its original action was faithful.

1 point: Derives the translate identity using normality.
1 point: Proves the block property, transitivity and equal sizes.
1 point: Proves containment of M in the block kernel.
1 point: Checks a genuine strict-containment counterexample and distinguishes the original kernel.

Targeted lesson returns:

Question 2: A large cyclic subgroup without a full classification

10 points

Let G be any group of order 105=3·5·7. You may use Sylow and Cauchy, but justify each count and normality assertion. Suggested time: 24 minutes, not learner-calibrated.

(a) · 3 points

Prove at least one Sylow 5-subgroup or Sylow 7-subgroup is normal.

(b) · 3 points

Show G contains a subgroup of order 35 in either case from part a. Prove every group of order 35 is cyclic.

(c) · 4 points

Prove any order-35 subgroup H of G is normal and G is a semidirect product C₃₅⋊C₃. Construct a nontrivial action showing the result need not be cyclic.

Review question 2: worked argument and scoring criteria

(a)

Sylow gives n₇=1 or 15, since it divides 15 and is 1 modulo 7; similarly n₅=1 or 21. Distinct subgroups of prime order intersect only at identity. If neither were normal, the order-seven subgroups would contribute 15·6=90 nonidentity elements and the order-five subgroups 21·4=84. These sets are disjoint by element order, giving 174>104 nonidentity elements, impossible. Therefore one count is 1, proving normality.

1 point: Obtains both exact alternatives for each Sylow count.
1 point: Justifies the element-count disjointness.
1 point: Concludes normality from the counting contradiction.

(b)

If P of order 7 is normal, a subgroup of order 5 in G/P, supplied by Cauchy, has preimage of order 35. If Q of order 5 is normal, use an order-seven subgroup in G/Q instead. In a group of order 35, Sylow forces both prime-order subgroups normal: n₇ divides 5 and is 1 modulo 7; n₅ divides 7 and is 1 modulo 5. Their intersection is trivial. A commutator belongs to both normal subgroups, hence is identity, so they commute. Their product is the whole group and the product of their generators has order 35 by coprimality.

1 point: Constructs the order-35 preimage in both cases.
1 point: Proves both Sylow subgroups of that group are normal.
1 point: Proves commutation and a generator of order 35.

(c)

The action on the three left cosets of H gives a transitive image in S₃. Its order divides both 105 and 6 and is divisible by 3, so it is 3. The point stabilizer in this image is trivial; its preimage is H, hence H is the action kernel and is normal. Cauchy supplies z of order 3; H∩⟨z⟩ is trivial and H⟨z⟩ has order 105. This is the internal semidirect product. Multiplication by 16 modulo 35 is an automorphism of C₃₅ of order 3: 16²≡11 and 16³≡1, while 16≠1. It defines a nontrivial C₃-action. In the external semidirect product the two chosen generators do not commute, so that group is not cyclic. No exhaustive isomorphism classification is asserted.

1 point: Uses the coset-action image to prove H is the kernel.
1 point: Constructs a complement and proves the internal product conditions.
1 point: Checks the unit and order of 16 modulo 35.
1 point: Proves the constructed action gives a noncyclic group and limits the claim.

Targeted lesson returns:

Question 3: Same subgroup and quotient, different actions

10 points

For u∈{4,16}, define Gᵤ on Z/17Z×Z/4Z by (a,b)(c,d)=(a+uᵇc,b+d). You may invoke the external semidirect-product theorem after checking its action hypothesis. Suggested time: 24 minutes, not learner-calibrated.

(a) · 3 points

Check both actions, give the inverse, and identify a normal C₁₇ with quotient C₄.

(b) · 3 points

Compute Z(G₄) and [G₄,G₄], with [x,y]=xyx⁻¹y⁻¹.

(c) · 4 points

Count the elements of each order in G₄. Compute Z(G₁₆) and decide whether the groups are isomorphic.

Review question 3: worked argument and scoring criteria

(a)

Both u are units. Modulo 17, 4²=−1, 4⁴=1 and 16²=1. Thus b↦(c↦uᵇc) is a well-defined homomorphism C₄→Aut(C₁₇), and the construction is a group. The inverse of (a,b) is (−u⁻ᵇa,−b), verified on both sides. The first-coordinate subgroup N={(a,0)} is cyclic of order 17 and is the kernel of the onto second-coordinate map to C₄, giving normality and the quotient.

1 point: Verifies the homomorphism condition, not just invertibility of u.
1 point: Checks the two-sided inverse.
1 point: Identifies the kernel and quotient through the surjective map.

(b)

Commutation with r=(1,0) forces 4ᵇ=1, hence b=0. Commutation with s=(0,1) then forces 4a=a, so a=0. Since r,s generate, the center is trivial. The quotient by N is abelian, so the commutator subgroup is contained in N. But [s,r]=(3,0), a generator of N; hence [G₄,G₄]=N.

1 point: Derives and solves both center restrictions.
1 point: Uses the abelian quotient for one containment.
1 point: Calculates a commutator generating the reverse containment.

(c)

For b=0 there is one identity and 16 elements of order 17. For b=2 the square is (a−a,0), so all 17 elements have order 2. For b=1 or 3 the second coordinate has order 4 and the first coordinate of the fourth power is a(1+4ᵇ+4²ᵇ+4³ᵇ)=0, giving 34 elements of exact order 4. The counts total 68. In G₁₆, commuting with r requires b even and commuting with s requires a=−a, hence a=0. Its center is {(0,0),(0,2)}. Unequal center orders rule out an isomorphism despite the common normal subgroup and quotient types: the action is additional data.

1 point: Counts the b=0 and b=2 orders correctly.
1 point: Proves the odd-b orders and checks the total.
1 point: Computes the exact second center.
1 point: Uses a valid invariant and explains the missing action data.

Targeted lesson returns:

Question 4: Decode a presentation and count its quotient maps

10 points

Let A=Z²/⟨(8,0),(0,12),(6,6)⟩, with generator images a,b. Thus 8a=12b=6a+6b=0. Establish the full group, not only an order bound. Suggested time: 24 minutes, not learner-calibrated.

(a) · 3 points

Show 4a=0. Put c=a−3b and prove A≅C₂×C₁₂, with c,b corresponding to the standard generators. Give maps both ways.

(b) · 3 points

Find the orders of a,b,c and the number of elements of exact order 6.

(c) · 4 points

Count all homomorphisms and all surjections A→C₆. Prove exactly three kernels occur among the surjections.

Review question 4: worked argument and scoring criteria

(a)

Twice the third relation minus 12b=0 yields 12a=0; subtract 8a=0 to obtain 4a=0. Thus 2a+6b=0, and 2c=2a−6b=0 because 12b=0. Sending the product generators to c,b defines a surjection since a=c+3b. Conversely send a to (1,3) and b to (0,1). The original relations map to (0,24),(0,12),(0,24), all zero. This reverse map sends c to (1,0). The two maps are inverse on generators and hence everywhere.

1 point: Derives 4a=0 and 2c=0.
1 point: Constructs the onto map from the proposed product.
1 point: Checks all relations for the reverse map and proves mutual inversion.

(b)

Their coordinates give orders lcm(2,4)=4, 12 and 2 respectively. The number killed by k in C₂×C₁₂ is gcd(2,k)gcd(12,k), by solving kx=0 in each cyclic factor. There are 12 killed by 6, 4 killed by 2 and 3 killed by 3; the latter sets intersect only at identity. Thus exactly 12−4−3+1=6 elements have order 6.

1 point: Computes all three generator orders.
1 point: Justifies the killed-by-k count.
1 point: Uses correct inclusion–exclusion for exact order.

(c)

The image of c is 0 or 3 and that of b is arbitrary, giving 12 maps. If b maps to 1 or 5, either c choice works: four surjections. If b maps to 2 or 4, only c↦3 works: two more. Images 0 or 3 for b never generate C₆ together with c. Thus there are six surjections. Two surjections have the same kernel iff their induced isomorphisms from A/K differ by a codomain automorphism. Aut(C₆) has two elements, multiplication by 1 or 5, and acts freely on surjections because an automorphism fixing an onto map is identity. Each kernel therefore occurs twice: three kernels, each of order 24/6=4.

1 point: Counts all maps from generator-image constraints.
1 point: Exhausts the surjective choices.
1 point: Proves the same-kernel/automorphism equivalence.
1 point: Checks freeness and obtains three kernels.

Targeted lesson returns:

2. Representations and geometry

30 practice points.

Question 5: Recover a hidden representation from incidence data

10 points

S₄ permutes the orthonormal basis eᵢⱼ of V=C⁶ indexed by unordered pairs. Its classes have representatives 1,(12),(12)(34),(123),(1234) and sizes 1,6,3,8,6. You may use Maschke, Schur and character orthogonality, including the norm-one irreducibility criterion. The inner product is ⟨χ,ψ⟩=(1/24)Σgχ(g)conjugate(ψ(g)). Suggested time: 24 minutes, not learner-calibrated.

(a) · 3 points

Compute χV. For the sum-zero subspace W of the natural C⁴ permutation representation, compute χW and prove irreducibility using its norm.

(b) · 3 points

Define T:C⁴→V by T(fᵢ)=Σj≠i eᵢⱼ. Prove T injective and equivariant. Construct a decomposition V=1⊕W⊕U and find χU.

(c) · 4 points

Prove U irreducible, determine its kernel, and compute dimC EndS₄(V). Justify the trace criterion you use for the kernel.

Review question 5: worked argument and scoring criteria

(a)

Trace counts fixed basis labels. Fixed-pair counts give χV=(6,2,2,0,0): a transposition fixes its own and the complementary pair; a double transposition fixes its two component pairs; neither remaining type fixes a pair. The four-letter character is (4,2,0,1,0); subtracting the constant line gives χW=(3,1,−1,0,−1). Its squared norm is (9+6+3+6)/24=1, so this actual nonzero representation is irreducible.

1 point: Derives all five fixed-pair counts.
1 point: Splits the natural constant line to obtain χW.
1 point: Uses the correct class weights and irreducibility criterion.

(b)

Relabeling vertices commutes with incidence, so T is equivariant. A kernel vector with coefficients aᵢ satisfies aᵢ+aⱼ=0 for every pair. Using any three indices forces their coefficients to vanish over C, and then the fourth vanishes. Thus T is injective. The constant line maps to 2Σeᵢⱼ and the sum-zero space maps isomorphically to W. The orthogonal complement U of im T is invariant because the permutation inner product is invariant, and has dimension 2. Therefore χU=χV−1−χW=(2,0,2,−1,0).

1 point: Proves equivariance on basis elements.
1 point: Proves injectivity from the pair coefficients.
1 point: Constructs the invariant complement and computes its character.

(c)

The squared norm of χU is (4+3·4+8)/24=1. A finite-order unitary matrix has eigenvalues of modulus one; if its trace equals its dimension, the sum of their real parts attains the maximum, forcing all eigenvalues 1. Unitary diagonalizability then makes the matrix identity. Thus χU(g)=2 exactly for kernel elements, namely identity and the three double transpositions. The constituents 1,W,U are pairwise inequivalent since their dimensions differ, each with multiplicity one. Schur gives scalar endomorphisms on each and no intertwiners between them, so the equivariant endomorphism space has dimension 3.

1 point: Checks the weighted norm of χU.
1 point: Proves trace equals dimension detects identity under the unitary hypothesis.
1 point: Identifies the exact kernel from the character.
1 point: Uses multiplicities and Schur to obtain dimension 3.

Targeted lesson returns:

Question 6: Tensor products with actual invariant subspaces

10 points

Let D₅=⟨r,s | r⁵=s²=1, srs=r⁻¹⟩ have order 10. On V=C² set r=diag(ζ,ζ⁻¹), where ζ=exp(2πi/5), and let s swap e₊,e₋. On V⊗V use g(v⊗w)=gv⊗gw. A bilinear form is complex-linear in both variables, not Hermitian. Suggested time: 24 minutes, not learner-calibrated.

(a) · 3 points

Verify the representation and prove it irreducible and faithful.

(b) · 3 points

Determine all invariant complex bilinear forms on V and which are nondegenerate. Why do these forms not imply an invariant line?

(c) · 4 points

Decompose V⊗V into irreducibles with explicit bases. Identify the one-dimensional characters and prove the two-dimensional constituent is not equivalent to V.

Review question 6: worked argument and scoring criteria

(a)

The matrices satisfy r⁵=s²=I, and swapping the two diagonal entries gives srs=r⁻¹. An invariant line must be an eigenline of r because its eigenvalues are distinct, but s exchanges those two lines; hence V is irreducible. Among rᵏ only k=0 gives identity, and every rᵏs is off-diagonal and not identity. These normal forms exhaust the ten elements, so the representation is faithful.

1 point: Checks every defining relation.
1 point: Rules out invariant lines using both generators.
1 point: Checks faithfulness on both normal-form types.

(b)

Write the matrix M=[[a,b],[c,d]]. The equation rᵀMr=M forces a=d=0 because ζ² and ζ⁻² are not 1. The equation sᵀMs=M forces b=c. Thus M=b[[0,1],[1,0]], nondegenerate exactly when b≠0 since det M=−b². A nondegenerate invariant form identifies V with its dual; it does not produce an invariant one-dimensional subspace. Its two isotropic coordinate lines are exchanged by s, consistent with irreducibility.

1 point: Solves r-invariance using the bilinear transpose convention.
1 point: Imposes s-invariance and proves the nondegeneracy condition.
1 point: Distinguishes self-duality from an invariant line.

(c)

The lines generated by e₊⊗e₋+e₋⊗e₊ and e₊⊗e₋−e₋⊗e₊ are fixed by r, and s acts by 1 and −1 respectively. They are the trivial and reflection-sign representations. The remaining space Z=span(e₊⊗e₊,e₋⊗e₋) is invariant with r=diag(ζ²,ζ⁻²) and s swapping. These four displayed vectors form a basis, since the mixed tensors are recovered by division by 2. The distinct-eigenline argument again proves Z irreducible. The r-eigenvalue exponents for V are {1,4} modulo 5 and for Z are {2,3}; equivalence would preserve the eigenvalue set of r, so they are inequivalent.

1 point: Finds and checks both invariant lines and their characters.
1 point: Constructs the two-dimensional block and proves the decomposition direct and exhaustive.
1 point: Proves irreducibility of the remaining block.
1 point: Uses a conjugacy invariant to prove inequivalence.

Targeted lesson returns:

Question 7: Find the center before naming a rotation group

10 points

On R³ let F(x,y,z)=(z+2,x+1,y−3) and H(x,y,z)=(x,4−y,−2−z). Let Γ be the group they generate. An affine isometry has form Qx+b with Q orthogonal. Prove finiteness before using an averaging argument. Suggested time: 24 minutes, not learner-calibrated.

(a) · 3 points

Find the unique common fixed point c. Translate c to the origin and prove the resulting matrices are orientation-preserving orthogonal transformations.

(b) · 3 points

Prove the translated group has exactly 12 elements and identifies with A₄ on the vertices d₁=(1,1,1), d₂=(1,−1,−1), d₃=(−1,1,−1), d₄=(−1,−1,1).

(c) · 4 points

Compute the average of the orbit of c+(1,0,0). Prove any finite group of affine transformations of a real vector space has a common fixed point by orbit averaging. Give counterexamples to uniqueness and to the converse claim that a fixed point forces the group finite.

Review question 7: worked argument and scoring criteria

(a)

H fixes precisely points with y=2,z=−1. F then forces x=z+2=1, giving c=(1,2,−1), which both maps fix. The translated maps are R(v₁,v₂,v₃)=(v₃,v₁,v₂) and D=diag(1,−1,−1). Both have orthonormal columns. The three-cycle matrix R and the matrix D with two negative entries both have determinant 1.

1 point: Solves, verifies and proves uniqueness of the common fixed point.
1 point: Computes both translated linear maps.
1 point: Checks orthogonality and determinant 1.

(b)

Conjugating D by R gives the other two sign matrices with exactly two negative entries. Together with identity they form a subgroup K of order 4 normalized by R. Its intersection with ⟨R⟩ is trivial and K⟨R⟩ is closed, so the generated group has exactly 4·3=12 elements. It preserves the displayed vertex set; D acts as (12)(34) and R as (234), both even. An element fixing all four vertices is identity because three of them span R³. Thus the permutation action is faithful with image a subgroup of A₄ of order 12, hence all A₄.

1 point: Constructs the sign subgroup and proves the exact group order.
1 point: Computes the vertex permutations and parity.
1 point: Proves faithfulness and identifies the image by order.

(c)

The translated orbit is {±e₁,±e₂,±e₃}, averaging to zero, so the requested average is c. Generally let O be a finite orbit and m=Σv∈O v/|O|. For T(v)=Qv+b, affine linearity gives T(m)=Σv∈O T(v)/|O|. Since T permutes O, this equals m. Uniqueness fails for {I,diag(1,−1)} on R², which fixes an entire line. A fixed point does not force finiteness: all planar rotations fix 0 and form an infinite group. Thus finiteness in this problem came from part b, and uniqueness from part a, not from averaging.

1 point: Checks the six-point orbit and its average.
1 point: Proves the affine averaging identity and fixed-point result.
1 point: Checks a finite-group counterexample to uniqueness.
1 point: Checks an infinite fixed-point group and distinguishes the logical directions.

Targeted lesson returns:

3. Synthesis and counterexamples

30 practice points.

Question 8: Representation degrees from a multiplication rule

10 points

Let H consist of triples (a,b,c)∈F₃³ with (a,b,c)(u,v,w)=(a+u,b+v,c+w+av). You may use that the number of complex irreducibles equals the number of conjugacy classes and that their squared degrees sum to |H|. Prove the other structural claims. Suggested time: 24 minutes, not learner-calibrated.

(a) · 3 points

Verify the group axioms and prove every nonidentity element has order 3.

(b) · 3 points

Find the center, commutator subgroup and number of conjugacy classes, using [x,y]=xyx⁻¹y⁻¹.

(c) · 4 points

Determine all irreducible complex representation degrees. Compare with C₃³, which has the same order and element orders.

Review question 8: worked argument and scoring criteria

(a)

For a third triple (r,s,z), the third coordinate of either association is c+w+z+av+as+us; the first two coordinates associate by addition. Identity is (0,0,0), and the inverse is (−a,−b,−c+ab), checked on both sides. Induction gives (a,b,c)ᵐ=(ma,mb,mc+binom(m,2)ab). At m=3 each coordinate is zero in F₃. Therefore a nonidentity element has exact order 3.

1 point: Checks associativity including the third coordinate.
1 point: Checks identity and two-sided inverse.
1 point: Computes the cube or power formula and deduces exact order.

(b)

The commutator is (0,0,av−ub). To commute with every (u,v,w), one needs a=b=0, so the center Z has three elements. Every commutator lies in Z, while [(1,0,0),(0,1,0)]=(0,0,1) generates it; hence [H,H]=Z. For a noncentral element, av−ub=0 is one nonzero linear equation in (u,v), with three solutions; w is arbitrary. Its centralizer has order 9, so its conjugacy class has size 3. There are three central singleton classes and 24/3=8 others, giving 11 classes.

1 point: Calculates the commutator and exact center.
1 point: Proves both commutator-subgroup containments.
1 point: Justifies the centralizer sizes and complete class count.

(c)

One-dimensional characters kill commutators and factor through H/Z≅C₃²; conversely every character of this quotient lifts. Choosing the images of two generators independently among the cube roots of unity gives exactly nine such characters. There are 11 irreducibles, so two further integer degrees d,e≥2 satisfy d²+e²=27−9=18. Neither can be at least 4, as then the sum is at least 20; among 2 and 3 only 3,3 work. Thus there are nine degree-one and two degree-three irreducibles. For C₃³ every irreducible is one-dimensional by Schur, and its three generators give 27 characters. H is nonabelian by the computed commutator; equal group order and equal element-order statistics therefore do not classify groups.

1 point: Proves one-dimensional characters factor through the abelianization in both directions.
1 point: Counts the nine characters explicitly.
1 point: Uses both class count and dimension squares to exhaust the other degrees.
1 point: Checks the abelian comparison and explains the failed classification criterion.

Targeted lesson returns:

Question 9: A permutation module that cannot be split

10 points

Over F₃ let V=F₃³ and let a generator of C₃ act by T(x,y,z)=(z,x,y). Put N=T−I, L=span(1,1,1), and U={vectors whose coordinate sum is zero}. Compare with the same permutation matrix over C. A decomposition must have invariant summands. Suggested time: 24 minutes, not learner-calibrated.

(a) · 2 points

Prove 0⊊L⊊U⊊V is invariant and all its successive one-dimensional quotients are trivial representations.

(b) · 4 points

Determine every endomorphism commuting with T and prove its only idempotents are 0 and I. Deduce no decomposition into two nonzero invariant summands exists.

(c) · 4 points

Over C give three explicit invariant lines decomposing C³. Explain precisely why averaging a projection works there but not over F₃.

Review question 9: worked argument and scoring criteria

(a)

Constants are fixed and coordinate sum is preserved. In characteristic 3 the constant vector has sum zero, so L⊆U; dimensions 1 and 2 give strictness. V/U is trivial by the invariant coordinate-sum map. For (x,y,z)∈U, N(x,y,z)=(z−x,x−y,y−z) has three equal coordinates in F₃, since x+y+z=0. Thus N(U)⊆L and U/L is trivial. T is already identity on L.

1 point: Checks invariance, strict dimensions and L⊆U in characteristic three.
1 point: Proves triviality of all three successive quotients.

(b)

The vectors e₁,Te₁,T²e₁ form a basis. A commuting map is determined by its arbitrary value on e₁, hence is uniquely a linear combination of I,T,T², equivalently I,N,N². In characteristic 3, N³=T³−I=0, and this polynomial change of basis preserves independence. For E=aI+bN+cN², the equations E²=E are a²=a, (2a−1)b=0, (2a−1)c+b²=0. The field equation gives a=0 or 1, so 2a−1 is nonzero and b=c=0. Only 0 and I are idempotent. An invariant decomposition would give a projection commuting with T that is neither 0 nor I, contradiction. Trivial flag quotients need not split as a direct sum.

1 point: Uses the cyclic basis to exhaust the commuting maps.
1 point: Justifies the nilpotent basis and all idempotent equations.
1 point: Solves the equations correctly over F₃.
1 point: Connects invariant decomposition to an equivariant projection.

(c)

For λ∈{1,ζ,ζ²}, ζ=exp(2πi/3), vλ=(1,λ⁻¹,λ⁻²) satisfies Tvλ=λvλ. Eigenvectors of distinct eigenvalues are independent: successive factors T−λI isolate a coefficient in any dependence. Thus these three lines give C³. Over C, for a projection P onto an invariant subspace, E=(1/3)Σj=0² TʲPT⁻ʲ commutes with T, has image in the subspace and restricts to identity there. It is therefore a projection with invariant kernel, yielding a complement. Over F₃, 1/3 does not exist; the unnormalized sum restricts to 3I=0, not identity. Part b proves an actual failure of splitting, not merely a failure of this particular averaging proof.

1 point: Gives and checks the three eigenvectors.
1 point: Justifies independence and the full decomposition.
1 point: Proves why the averaged map is an equivariant projection over C.
1 point: Identifies the missing scalar and relates it to the actual counterexample.

Targeted lesson returns:

Question 10: Repair a false direct-product certificate

10 points

Let G=(C₄×C₄)/⟨(2,2)⟩, additively, and let H,K be the images of its two coordinate factors. The quotient subgroup has two elements. Audit the claim: H and K are normal, commute and generate, therefore G≅H×K. Suggested time: 24 minutes, not learner-calibrated.

(a) · 3 points

Find H,K,H∩K and prove G≅C₄×C₂ by an explicit map.

(b) · 3 points

If subgroups H,K commute elementwise and G=HK, prove the multiplication map H×K→G is a surjective homomorphism with kernel {(z,z⁻¹):z∈H∩K}. Apply the correction to this example.

(c) · 4 points

Prove the direct-product claim becomes valid when H,K are finite normal subgroups of coprime orders and G=HK, without assuming commutation. Contrast an overlap failure with a nontrivial-action failure.

Review question 10: worked argument and scoring criteria

(a)

Neither coordinate factor meets ⟨(2,2)⟩ nontrivially, so each image is C₄. Equality of the cosets of (a,0) and (0,b) means (a,−b) is (0,0) or (2,2). The intersection therefore consists of identity and the common coset of (2,0)=(0,2). The map (a,b)↦(a+b mod 4,b mod 2) onto C₄×C₂ is surjective: choose b of the requested parity and then a. Its kernel is exactly {(0,0),(2,2)}, so it induces the claimed quotient isomorphism; G has order 8.

1 point: Checks that both factor images retain order 4.
1 point: Computes the full intersection from the quotient relation.
1 point: Proves the explicit map onto with exactly the required kernel.

(b)

Commutation gives (h₁k₁)(h₂k₂)=h₁h₂k₁k₂, proving the homomorphism property; G=HK proves surjectivity. The product hk equals identity exactly when h=k⁻¹, which is equivalent to (h,k)=(z,z⁻¹) for z in the intersection. Thus G is the quotient by this kernel, not automatically the direct product. Here a product of order 16 is divided by the two-element kernel to obtain order 8. Normality and commutation do not remove the overlap.

1 point: Checks the homomorphism and onto properties with their hypotheses.
1 point: Proves the exact kernel in both directions.
1 point: States the quotient correction and applies its orders.

(c)

For h∈H,k∈K the commutator hkh⁻¹k⁻¹ belongs to K by its normality and to H by its normality. The intersection has order dividing both coprime orders, so is trivial; all these commutators are identity. Part b then gives the direct product. The displayed quotient is an overlap failure: the subgroups commute but intersect nontrivially. For an action failure take S₃ with H=A₃ and K=⟨(12)⟩. They have trivial intersection and generate, but K is not normal and conjugation by (12) inverts the three-cycle. The multiplication map is a bijection of sets, not a direct-product homomorphism; S₃ is nonabelian while C₃×C₂ is abelian.

1 point: Places the commutator in both normal subgroups.
1 point: Uses coprime orders and the multiplication map to prove the direct product.
1 point: Explains the checked overlap counterexample.
1 point: Checks the S₃ action counterexample and names the failed hypothesis.

Targeted lesson returns:

Turn the review into a new argument

Award points for reasoning actually present. Allow valid later reasoning based on a carried-forward value unless its criterion requires the correct value; do not award correct-result points for an incorrect result. Record missing hypotheses, unjustified inverses, invalid identifications and arithmetic errors separately.

The total is out of 100, with no university-grade conversion. Preserve your first attempt, consult the targeted lesson, explain the repair and attempt a fresh case. A teacher or peer may review the complete argument; if nobody did, label it self-review. Public worked solutions cannot certify independent proficiency.

Return to Abstract algebra I: groups and symmetry for more focused practice. For sustained synthesis, build a presentation-and-operator manuscript with exact certificates, revisions and a separate changed case.

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