MATH 401, EXAM 1

DANIEL R. GRAYSON

Problem 1. True or false? (No support needed for your answer.)

(a) The group ℤ/7ℤ × ℤ/7ℤ is cyclic. [3 points] False, since there is no element of order 49.

(b) Finite abelian groups are solvable. [3 points] True, since the factors in the composition series of an abelian group are abelian.

(c) The group S_{5} is solvable. [3 points] False, since the nonabelian group A_{5} is one of the factors in its composition series.

(d) Groups of prime order are simple. [3 points] True, since the order of a subgroup divides the order of the group.

(e) The group ℤ/2ℤ × ℤ/7ℤ is cyclic. [3 points] True, since (1, 1) has order 14.

(f) The only subgroup of order 60 in S_{5} is A_{5}. [7 points] True, since any other subgroup of the same order would be normal (the only other coset is both a left coset and a right coset), and the intersection with the simple subgroup A_{5} would be a normal subgroup of it.

(g) The centralizer of (12)(34) in A_{5} is cyclic. [8 points] False, since the centralizer is {1, (12)(34), (13)(24), (14)(23)}, and is isomorphic to ℤ/2ℤ × ℤ/2ℤ. Problem 2. Consider the multiplicative group G of matrices g with entries in the

field 𝔽_{2} = ℤ/2ℤ of the form g = (\begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix}) .

(a) Show G is generated by the elements p = (\begin{matrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) and q = (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}) .

[15 points] We can compute a product like pg most easily by performing the appropriate row operation on g, namely, adding row 2 to row 1. Similarly, qg is obtained from g by adding row 3 to row 2. We see that p^{2} = q^{2} = 1, so we

investigate words like r = pqpq = (\begin{matrix}1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}). The matrices p, q, r give us all the elementary row operations we need; explicitly, p^{a}q^{c}r^{b}g = 1. Alternatively, once you find that pq ≠ qp, we see that pqpq = [p, q] ≠ 1, and it follows that pq has order 4. Since p has order 2, and p ≠ pqpq, we have obtained at least 5 elements, namely, {p, 1, pq, (pq)^{2}, (pq)^{3}}. The order of the subgroup generated by p and q is at least 5 and is a divisor of 8, so must be 8.

Date: October 8, 1999.

(b) Let V = (𝔽_{2})^{3} and think of its elements as column vectors, so that G acts on V on the left as usual. Compute the set G\V.

[15 points] If v ∈ V has a nonzero entry in row i, then row operations are available in G to modify the entry in row j for any j < i, and there is no way to modify the entry in row i. Hence an orbit is completely determined by the lowest nonzero row of a vector in the orbit, and G\V = {{0}, {e⃗_{1}}, 𝔽_{2}e⃗_{1} + e⃗_{2}, 𝔽_{2}e⃗_{1} + 𝔽_{2}e⃗_{2} + e⃗_{3}}, where e⃗_{i} denotes the standard basis vector with a 1 just in row i.

(c) Let X ∈ G\V be the largest orbit and show the resulting homomorphism G → S_{X} is injective. Are there any blocks in X?

[10 points] The largest orbit is X = 𝔽_{2}e⃗_{1} + 𝔽_{2}e⃗_{2} + e⃗_{3}, and from ge⃗_{3} = (\begin{matrix}b \\ c \\ 1 \end{matrix}) and

g(e⃗_{2} + e⃗_{3}) = (\begin{matrix}a + b \\ 1 + c \\ 1 \end{matrix}) we can recover g, so the homomorphism is injective. Since

ge⃗_{1} = e⃗_{1} we see there are blocks {e⃗_{3}, e⃗_{1} + e⃗_{3}} and {e⃗_{2} + e⃗_{3}, e⃗_{1} + e⃗_{2} + e⃗_{3}} .

(d) Show G is isomorphic to a group we’ve seen before.

[10 points] Using each of the blocks found above as the pair of labels for diagonally opposed vertices of a square, we see that any g ∈ G gives a symmetry of the square, hence G ≅ D_{8}.

Problem 3. Let G be a finite group with a subgroup H whose index p = |G/H| is the smallest prime number dividing |G|. Show H is a normal subgroup of G. [Hint: Let G act on the set G/H and consider the resulting homomorphism to the symmetric group S_{G/H}. How does the kernel compare to H? What is the index of the kernel?]

[20 points] We let G act on the set G/H by left multiplication g · g′H = gg′H. It acts transitively. The image of the resulting homomorphism ϕ : G → S_{G/H} is a group whose order is a divisor both of |G| and of |S_{G/H}| = p!. The greatest common divisor of |G| and p! is p, according to the hypothesis, so the image has order p, and hence the kernel K of ϕ is a normal subgroup of G of index p. We see that K ⊆ H since k · H = H for k ∈ K. Hence K = H, since they have the same index, and hence H is normal.

University of Illinois at Urbana-Champaign

   E-mail address: dan@math.uiuc.edu
   URL: http://www.math.uiuc.edu/~dan