MATH 201, Algebra II, Midterm (February 5, 1999)

(Winter 1999, R. Boltje)

1. Let A be a finitely generated abelian torsion group with elementary divisors 2, 4, 4, 3, 3, 5, 7, 49. What are the invariant factors of A?

2. Determine the Jordan normal form of the matrix

A = (\begin{matrix}−1 & 0 & 0 \\ 1 & 0 & 1 \\ −1 & −1 & −2 \end{matrix}) .

3. Let R be a ring, let M and N be R-modules, and let f ∈ Hom_{R}(M, N), g ∈ Hom_{R}(N, M) such that g ∘ f = id_{M}. Show that N = im(f) ⊕ ker(g).

4. Let R be a ring and let M and N be R-modules. Consider the abelian group F := Hom_{R}(M, N). Does

(rf)(m) := r(f(m)), for r ∈ R, f ∈ F, m ∈ M,

define an R-module structure on F? (Give a proof or a counterexample.) If the answer is “no”, try to impose a condition on R such that the answer is “yes”.

5. Let G be a finite abelian group of order n. Show that for each divisor m of n there exists a subgroup H of G of order m.

Solutions

1. The invariant factors of A are d_{3} = 4 · 3 · 5 · 49, d_{2} = 4 · 3 · 7, d_{1} = 2, since d_{1} | d_{2} | d_{3} and the prime power factors of d_{1}, d_{2}, d_{3} are the given numbers.

2. We transform the matrix λ1 − A using the elementary row and column operations:

(\begin{matrix}λ + 1 & 0 & 0 \\ −1 & λ & −1 \\ 1 & 1 & λ + 2 \end{matrix}) → (\begin{matrix}1 & 1 & λ + 2 \\ −1 & λ & −1 \\ λ + 1 & 0 & 0 \end{matrix})

→ (\begin{matrix}1 & 1 & λ + 2 \\ 0 & λ + 1 & λ + 1 \\ λ + 1 & 0 & 0 \end{matrix}) → (\begin{matrix}1 & 1 & λ + 2 \\ 0 & λ + 1 & λ + 1 \\ 0 & −λ − 1 & −λ^{2} − 3λ − 2 \end{matrix})

→ (\begin{matrix}1 & 0 & λ + 2 \\ 0 & λ + 1 & λ + 1 \\ 0 & −λ − 1 & −λ^{2} − 3λ − 2 \end{matrix}) → (\begin{matrix}1 & 0 & 0 \\ 0 & λ + 1 & λ + 1 \\ 0 & −λ − 1 & −λ^{2} − 3λ − 2 \end{matrix})

→ (\begin{matrix}1 & 0 & 0 \\ 0 & λ + 1 & λ + 1 \\ 0 & 0 & −λ^{2} − 2λ − 1 \end{matrix}) → (\begin{matrix}1 & 0 & 0 \\ 0 & λ + 1 & 0 \\ 0 & 0 & −(λ + 1)^{2} \end{matrix})

→ (\begin{matrix}1 & 0 & 0 \\ 0 & λ + 1 & 0 \\ 0 & 0 & (λ + 1)^{2} \end{matrix}) .

Therefore, the Jordan normal form of A consists of 2 Jordan blocks, one of size 1, one of size 2, both with −1’s on the diagonal:

(\begin{matrix}−1 & 0 & 0 \\ 0 & −1 & 0 \\ 0 & 1 & −1 \end{matrix}) .

3. Let n ∈ N, then n = f(g(n)) + [n − f(g(n))] with f(g(n)) ∈ im(f) and n − f(g(n)) ∈ ker(g), since g(n) − g(f(g(n))) = g(n) − g(n) = 0. Moreover, if n ∈ im(f) ∩ ker(g), then there exists m ∈ M with n = f(m). But since n ∈ ker(g), we have m = g(f(m)) = g(n) = 0, and so also n = f(m) = 0.

4. First note that for f ∈ F, r ∈ R and m ∈ M we have r(f(m)) = f(rm), since f is R-linear. The definition in the text does not give an Rmodule structure on F, since rf is in general not again in F. In fact, we need to show that (rf)(sm) = s((rf)(m)) for all r, s ∈ R, f ∈ F, and m ∈ M. The left hand side is now given by

(rf)(sm) = f(r(sm)) = f((rs)m) = (rs)(f(m)) ,

and the right hand side is given by

s((rf)(m)) = s(f(rm)) = s(r(f(m))) = (sr)(f(m)) .

If we take M = N = R for a non-commutative ring R and if we take r, s ∈ R with rs ≠ sr, then with f = id_{R} and m = 1 the above equation does not hold. But it does hold, if R is commutative. Finally, we check that (r +s)f = rf +sf, r(f + g) = rf + fg, and 1f = f for all r, s ∈ R and f, g ∈ F. Let m ∈ M, then

((r + s)f)(m) = f((r + s)m) = f(rm + sm)

= f(rm) + f(sm) = (rf)(m) + (sf)(m) = (rf + sf)(m) ,

(r(f + g))(m) = (f + g)(rm) = f(rm) + g(rm)
= (rf)(m) + (rg)(m) = (rf + rg)(m) ,
(1f)(m) = f(1m) = f(m) .

5. We proof the assertion by induction on the number of primes divisors in a decomposition of n/m. If n/m = 1, we can take G = H. Assume next that n/m is a prime p. We write G ≅ ℤ/p_{1}^{α_{1}}ℤ ⊕ ⋯ ⊕ ℤ/p_{t}^{α_{t}}ℤ with prime numbers p_{1}, …, p_{t}. Then one of the p_{i}, say p_{1} is equal to p, and the subgroup H of G corresponding to pℤ/p_{1}^{α_{1}}ℤ ⊕ ⋯ ⊕ ℤ/p_{t}^{α_{t}}ℤ has index p in H, since (ℤ/p_{1}^{α_{1}}ℤ)/(pℤ/p_{1}^{α_{1}}ℤ) ≅ ℤ/pℤ. Therefore, |H| = |G|/[G : H] = n/p = m. Now assume that n/m = p_{1} ⋯ p_{k} is a product of k primes with k > 1. By induction there exists a subgroup U of G of order mp_{k}, and since U is again an abelian torsion group, we can use the argument for the case k = 1 again and obtain a subgroup H of U of index p_{k} which then has order m.