Tag Archives: abelian group

The set of ideal classes over an algebraic intger ring is an abelian group

Let \mathcal{O} be the set of integers in an algebraic number field K, and let G = \mathcal{O}/\sim denote the set of equivalence classes of ideals in \mathcal{O} under the relation A \sim B if and only if (\alpha)A = (\beta)B for some nonzero \alpha,\beta \in \mathcal{O}. Prove that G is an abelian group under the usual ideal product of class representatives.


We will denote by [A] the equivalence class containing the ideal A. We showed in this previous exercise that [A][B] = [AB] is a well-defined binary operator on G.

For all [A], [B], [C] \in G, we have [A]([B][C]) = [A][BC] = [A(BC)] = [(AB)C] = [AB][C] ([A][B])[C], so that our product is associative.

For all [A], we have [A][(1)] = [A(1)] = [A], and similarly [(1)][A] = [A].

For all [A], there exists (by Theorem 8.13) an ideal B such that AB is principal. Thus [A][B] = [(1)], and so every element of G has an inverse.

Hence G is a group.

Moreover, we have [A][B] = [AB] = [BA] = [B][A], so that G is abelian.

The injective hull of the ZZ-module ZZ is QQ

Prove that the injective hull of the \mathbb{Z}-module \mathbb{Z} is \mathbb{Q}.


Recall that an injective hull of a module M is an injective module Q such that there exists an injective homomorphism M \rightarrow Q and such that every injective module homomorphism from M to an injective module A lifts to an injective homomorphism from Q to A. We will take it as given that injective hulls exist and are unique up to isomorphism.

Let H be the injective hull of \mathbb{Z} as a \mathbb{Z}-module. Since \mathbb{Q} is injective as a \mathbb{Z}-module and the inclusion \mathbb{Z} \rightarrow \mathbb{Q} is injective, there exists an injective abelian group homomorphism H \rightarrow \mathbb{Q}. We will identify H with its image in \mathbb{Q}.

Note that H is divisible as an abelian group, and since \mathbb{Z} \subseteq H, for all integers n \neq 0 there exists q \in H such that nq = 1. In particular, q = \frac{1}{n} \in H for all n \neq 0. Since H \subseteq \mathbb{Q} is a \mathbb{Z}-submodule, we have \frac{a}{b} \in H for all a,b \in \mathbb{Z} with b \neq 0. Thus H = \mathbb{Q}.

If Q is an injective R-module, then Hom(R,Q) over ZZ is an injective R-module

Let R be a ring with 1 and let M be a left unital R-module.

  1. Show that \mathsf{Hom}_\mathbb{Z}(R,M) is a left R-module via the action (r\varphi)(a) = \varphi(ar).
  2. Suppose \psi : A \rightarrow B is an injective R-module homomorphism. Prove that if every homomorphism \varphi : A \rightarrow M lifts to a homomorphism \Phi : B \rightarrow M such that \varphi = \Phi \circ \psi, then every homomorphism \theta : A \rightarrow \mathsf{Hom}_\mathbb{Z}(R,M) lifts to a homomorphism \Theta : B \rightarrow \mathsf{Hom}_\mathbb{Z}(R,M) such that \theta = \Theta \circ \psi.
  3. Prove that if Q is an injective R-module, then \mathsf{Hom}_\mathbb{Z}(R,Q) is also an injective R-module.

  1. Note that R is a (\mathbb{Z},R)-bimodule and M is a left \mathbb{Z}-module. By this previous exercise, via the action (r\varphi)(a) = \varphi(ar), \mathsf{Hom}_\mathbb{Z}(R,M) is a left R-module.
  2. Let \psi : A \rightarrow B be an injective R-module homomorphism. Now suppose that for every homomorphism \varphi : A \rightarrow M, there exists a homomorphism \Phi : B \rightarrow M such that \varphi = \Phi \circ \psi. Let \theta : A \rightarrow \mathsf{Hom}_\mathbb{Z}(R,M) be a homomorphism. Now define \varphi : A \rightarrow M by \varphi(a) = \theta(a)(1); certainly \varphi is well-defined. We claim that \varphi is also a module homomorphism. To see this, let a,b \in A and r \in R. Now \varphi(a+rb) = \theta(a+rb)(1) = \theta(a)(1) + r\theta(b)(1) = \varphi(a) + r \varphi(b) as desired. By our hypothesis, there is a homomorphism \Phi : B \rightarrow M such that \varphi = \Phi \circ \psi. Define \Theta : B \rightarrow \mathsf{Hom}_\mathbb{Z}(R,M) by \Theta(b)(r) = \Phi(rb). To see that \Theta is well-defined, let x,y \in R and r \in \mathbb{Z}. Now \Theta(b)(x+ry) = \Phi((x+ry)b) = \Phi(xb + ryb) = \Phi(xb) + r\Phi(yb) = \Theta(b)(x) + r\Theta(b)(y). So \Phi(b) is indeed a \mathbb{Z}-module homomorphism, and thus \Theta is well defined. Next, we claim that \Theta is an R-module homomorphism. To that end, let x,y \in B and r \in R. For all a, \Theta(x+ry)(a) = \Phi(a(x+ry)) = \Phi(ax + ary) = \Phi(ax) + \Phi(ary) = \Theta(x)(a) + \Theta(y)(ar) = \Theta(x)(a) + (r\Theta)(y)(a) = (\Theta(x) + r\Theta(y))(a). Thus \Theta(x+ry) = \Theta(x) + r\Theta(y), so \Theta is a module homomorphism. Finally, we claim that \Theta \circ \psi = \theta. To see this, note that (\Theta \circ \psi)(a)(r) = \Theta(\psi(a))(r) = \Phi(r\psi(a)) = \Phi(\psi(ra)) = (\Phi \circ \psi)(ra) = \varphi(ra) = \theta(ra)(1) = (r\theta(a))(1) = \theta(a)(r). So \Theta \circ \psi = \theta.
  3. Suppose Q is an injective (left, unital) R-module. Then for every injective homomorphism \psi : A \rightarrow B, if \varphi : A \rightarrow Q is a homomorphism then there exists \Phi : B \rightarrow Q such that \varphi = \Phi \circ \psi. By part (b), then, every homomorphism \theta : A \rightarrow \mathsf{Hom}_\mathbb{Z}(R,Q) lifts to a homomorphism \Theta : B \rightarrow \mathsf{Hom}_\mathbb{Z}(R,Q). So \mathsf{Hom}_\mathbb{Z}(R,Q) is injective as an R-module.

Finite abelian groups are not naturally QQ-modules

Every finite abelian group M is naturally a \mathbb{Z}-module. Can this ring action be extended to make M a \mathbb{Q}-module?


Let M be a (multiplicative) finite abelian group of order k. Suppose further that M is a \mathbb{Q} module and that the action of \mathbb{Q} on M extends the natural action of \mathbb{Z} given by k \cdot m = m^k. Let x \in M be any nonidentity element. Now \frac{1}{k} \cdot x = y for some y \in M. then k \cdot (\frac{1}{k} \cdot x) = k \cdot y, and so x = y^k = 1, a contradiction. Thus no such module structure exists.

Compute the annihilator of a module

let M be the abelian group (I.e. Z-module) \mathbb{Z}/(24) \times \mathbb{Z}/(15) \times \mathbb{Z}/(50).

  1. Compute \mathsf{Ann}_\mathbb{Z}(M).
  2. Compute \mathsf{Ann}_M(2\mathbb{Z}).

  1. We claim that \mathsf{Ann}_\mathbb{Z}(M) = (600). (\subseteq) Let k be in the annihilator of M. In particular, k \cdot (\overline{1}, \overline{1}, \overline{1}) = (\overline{k}, \overline{k}, \overline{k}) = 0. Thus 24|k, 15|k, and 50|k, so that \mathsf{lcm}(24,15,50) = 600 divides k. Thus k \in (600). (\supseteq) Let 600t \in (600). Now let (\overline{a}, \overline{b}, \overline{c}) \in M. Note that 600t \cdot (\overline{a}, \overline{b}, \overline{c}) = (\overline{25 \cdot 24 \cdot a}, \overline{40 \cdot 15 \cdot b}, \overline{12 \cdot 50c}) = 0. So 600t annihilates M. Thus \mathsf{Ann}_\mathbb{Z}(M) = (600).
  2. We claim that \mathsf{Ann}_M((2)) = (12)/(24) \times (15)/(15) \times (25)/(50). (\subseteq) Let (\overline{a}, \overline{b}, \overline{c}) annihilate (2). In particular, 2 \cdot (\overline{a}, \overline{b}, \overline{c}) = (\overline{2a}, \overline{2b}, \overline{2c}) = 0. Then 2a \equiv 0 mod 24, 2b \equiv 0 mod 15, and 2c \equiv 0 mod 50. Thus a \equiv 0 mod 12, b \equiv 0 mod 15, and c \equiv 0 mod 25. So (\overline{a}, \overline{b}, \overline{c}) \in (12)/(24) \times (15)/(15) \times (25)/(50). (\supseteq) Suppose (\overline{12a}, \overline{0}, \overline{25c}) \in (12)/(24) \times (15)/(15) \times (25)/(50), and let 2t \in (2). Now 2t \cdot (\overline{12a}, \overline{0}, \overline{25c}) = (\overline{24ta}, \overline{0}, \overline{50tc}) = 0; thus (\overline{12a}, \overline{0}, \overline{25c}) annihilates (2). So \mathsf{Ann}_M(2\mathbb{Z}) = (12)/(24) \times (15)/(15) \times (25)/(50) \cong \mathbb{Z}/(2) \times \mathbb{Z}/(2).

Definition and universal property of the direct limit of commutative unital rings

In this exercise, we develop the concept of direct limit. Let I be a nonempty paritally ordered set, and let \{A_i\}_I be a collection of abelian groups. We suppose also that I is directed; that is, for all i,j \in I, there exists k \in I with i,j \leq k.

Suppose that for every pair of indices i,j \in I with i \leq j, there is a map \rho_{i,j} : A_i \rightarrow A_j such that the following hold: (1) \rho_{j,k} \circ \rho_{i,j} = \rho_{i,k} whenever i \leq j \leq k and (2) \rho_{i,i} = 1 for all i \in I.

Let B = \bigcup_I A_i \times \{i\} be the disjoint union of the A_i. Define a relation \sigma on B as follows: (a,i) \ \sigma\ (b,j) if and only if there exists k \in I such that i,j \leq k and \rho_{i,k}(a) = \rho_{j,k}(b).

  1. Show that \sigma is an equivalence relation on B. We define \varinjlim A_i = B/\sigma.
  2. Let [x]_\sigma denote the class of x in \varinjlim A_i and define \rho_i : A_i \rightarrow \varinjlim A_i by \rho_i(a) = [(a,i)]_\sigma. Show that if each \rho_{i,j} is injective, then \rho_i is also injective for all i.
  3. Assume that the \rho_{i,j} are all group homomorphisms. For [(a,i)]_\sigma,[(b,j)]_\sigma \in \varinjlim A_i, show that the operation [(a,i)]_\sigma + [(b,j)]_\sigma = [(\rho_{i,k}(a) + \rho_{j,k}(b),k)]_\sigma, where k is any upper bound of i and j, is well defined and makes \varinjlim A_i an abelian group. Deduce that the \rho_i are group homomorphisms.
  4. Prove that if all the A_i are commutative rings with 1 \neq 0 and all the \rho_{i,j} are unital ring homomorphisms (that is, they send 1 to 1), then \varinjlim A_i may likewise be given the structure of a commutative ring with 1 \neq 0 such that the \rho_i are all ring homomorphisms.
  5. Under the hypotheses of part (c), prove that \varinjlim A_i has the following universal property: if C is any abelian group such that for each i \in I there is a homomorphism \varphi_i : A_i \rightarrow C with \varphi_i = \varphi_j \circ \rho_{i,j} whenever i \leq j, then there is a unique homomorphism \varphi : A \rightarrow C such that \varphi \circ \rho_i = \varphi_i for all i.

  1. To show that \sigma is an equivalence, we need to verify that it is reflexive, symmetric, and transitive.
    1. (\sigma is reflexive) Let (a,i) \in B. Note that i \leq i, and that \rho_{i,i}(a) = a = \rho_{i,i}(a). Thus (a,i) \ \sigma\ (a,i), and hence \sigma is reflexive.
    2. (\sigma is symmetric) Suppose (a,i) \ \sigma\ (b,j). Then there exists k \geq i,j such that \rho_{i,k}(a) = \rho_{j,k}(b). Certainly \rho_{j,k}(b) = \rho_{i,k}(a), so that (b,j)\ \sigma\ (a,i). Thus \sigma is symmetric.
    3. (\sigma is transitive) Suppose (a,i)\ \sigma\ (b,j) and (b,j)\ \sigma\ (c,k). Then there exist \ell \geq i,j such that \rho_{i,\ell}(a) = \rho_{j,\ell}(b) and m \geq j,k such that \rho_{j,m}(b) = \rho_{k,m}(c). Since I is a directed poset, there exists t \in I such that t \geq \ell,m. Now \rho_{i,t}(a) = \rho_{\ell,t}(\rho_{i,\ell}(a)) = \rho_{\ell,t}(\rho_{j,\ell}(b)) = \rho_{j,t}(b) = \rho_{m,t}(\rho_{j,m}(b)) = \rho_{m,t}(\rho_{k,m}(c)) = \rho_{k,t}(c). Thus (a,i)\ \sigma\ (c,k), and hence \sigma is transitive.

    Thus \sigma is an equivalence relation.

  2. Suppose that the \rho_{i,j} are all injective. Choose i \in I, and let a,b \in A_i such that \rho_i(a) = \rho_i(b). Then we have [(a,i)]_\sigma = [(b,i)]_\sigma. That is, for some k \geq i, we have \rho_{i,k}(a) = \rho_{i,k}(b). Since \rho_{i,k} is injective, a = b. Thus \rho_i is injective.
  3. Suppose that the \rho_{i,j} are all group homomorphisms. First we show that + is well defined.

    Let [(a_1,i_1)]_\sigma = [(a_2,i_2)]_\sigma and [(b_1,j_1)]_\sigma = [(b_2,j_2)]_\sigma. Then there exists s \geq i_1,i_2 such that \rho_{i_1,s}(a_1) = \rho_{i_2,s}(a_2) and r \geq j_1,j_2 such that \rho_{j_1,r}(b_1) = \rho_{j_2,r}(b_2). Now (using the directedness of I) choose arbitrary k_1 \geq i_1,j_1 and k_2 \geq i_2,j_2. Finally, again choose t \geq k_1,k_2,r,s. Now note the following.

    \rho_{k_1,t}(\rho_{i_1,k_1}(a_1) + \rho_{j_1,k_1}(b_1))  =  \rho_{k_1,t}(\rho_{i_1,k_1}(a_1)) + \rho_{k_1,t}(\rho_{j_1,k_1}(b_1))
     =  \rho_{i_1,t}(a_1) + \rho_{j_1,t}(b_1)
     =  \rho_{s,t}(\rho_{i_1,s}(a_1)) + \rho_{r,t}(\rho_{j_1,r}(b_1))
     =  \rho_{s,t}(\rho_{i_2,s}(a_2)) + \rho_{r,t}(\rho_{j_2,r}(b_2))
     =  \rho_{i_2,t}(a_2) + \rho_{j_2,t}(b_2)
     =  \rho_{k_2,t}(\rho_{i_2,k_2}(a_2)) + \rho_{k_2,t}(\rho_{j_2,k_2}(b_2))
     =  \rho_{k_2,t}(\rho_{i_2,k_2}(a_2) + \rho_{j_2,k_2}(b_2))

    Thus (\rho_{i_1,k_1}(a_1) + \rho_{j_1,k_1}(b_1),k_1) \ \sigma\ (\rho_{i_2,k_2}(a_2) + \rho_{j_2,k_2}(b_2),k_2), and we have [(a_1,i_1)]_\sigma + [(b_1,j_1)]_\sigma = [(a_2,i_2)]_\sigma = [(b_2,j_2)]_\sigma. Thus + is well-defined.

    Next we show that (\varinjlim A_i, +) is an abelian group.

    1. (+ is associative) Let [(a,i)], [(b,j)], and [(c,k)] be in \varinjlim A_i, and let \ell \geq i,j, t \geq j,k, and m \geq \ell,t. Then we have the following.
      \left( [(a,i)] + [(b,j)] \right) + [(c,k)]  =  [\rho_{i,\ell}(a) + \rho_{j,\ell}(b)] + [(c,k)]
       =  [(\rho_{\ell,m}(\rho_{i,\ell}(a) + \rho_{j,\ell}(b)) + \rho_{k,m}(c),m)]
       =  [(\rho_{\ell,m}(\rho_{i,\ell}(a)) + \rho_{\ell,m}(\rho_{j,\ell}(b)) + \rho_{k,m}(c),m)]
       =  [(\rho_{i,m}(a) + \rho_{j,m}(b) + \rho_{k,m}(c),m)]
       =  [(\rho_{i,m}(a) + \rho_{t,m}(\rho_{j,t}(b)) + \rho_{t,m}(\rho_{j,t}(c)),m)]
       =  [(\rho_{i,m}(a) + \rho_{t,m}(\rho_{j,t}(b) + \rho_{k,t}(c)),m)]
       =  [(a,i)] + [(\rho_{j,t}(b) + \rho_{k,t}(c),t)]
       =  [(a,i)] + \left( [(b,j)] + [(c,k)] \right)

      So + is associative.

    2. Note that for all i,j \in I, there exists k \geq i,j, and \rho_{i,k}(0) = \rho_{j,k}(0) since the \rho_{i,j} are group homomorphisms. Thus [(0,i)] = [(0,j)] for all i,j. Let 0 = [(0,i)]. Now let [(a,i)] \in \varinjlim A_i. Then 0 + [(a,i)] = [(0,i)] + [(a,i)] = [(\rho_{i,i}(0) + \rho_{i,i}(0), i)] = [(0+a,i)] = [(a,i)]. Similarly, [(a,i)] + 0 = [(a,i)]. Thus 0 = [(0,i)] is an additive identity element.
    3. Let [(a,i)] \in \varinjlim A_i. Note that [(a,i)] + [(-a,i)] = [(\rho_{i,i}(a) + \rho_{i,i}(-a),i)] = [(a-a,i)] = [(0,i)] = 0. Thus every element of \varinjlim A_i has an additive inverse.
    4. Let [(a,i)], [(b,j)] \in \varinjlim A_i, and let k \geq i,j. Then [(a,i)] + [(b,j)] = [(\rho_{i,k}(a) + \rho_{j,k}(b),k)] = [(\rho_{j,k}(b) + \rho_{i,k}(a),k)] = [(b,j)] + [(a,i)]. Thus + is commutative.

    Thus (\varinjlim A_i, +) is an abelian group. Finally, we show that each \rho_i : A_i \rightarrow \varinjlim A_i is a group homomorphism. Let a,b \in A_i. Then \rho_i(a+b) = [(a+b,i)] = [(\rho_{i,i}(a) + \rho_{i,i}(b),i)] = [(a,i)] + [(b,i)] = \rho_i(a) + \rho_i(b). Thus \rho_i is a group homomorphism for all i.

  4. Define an operator on \varinjlim A_i as follows: [(a,i)] \cdot [(b,j)] = [(\rho_{i,k}(a) \cdot \rho_{j,k}(b), k)], where k is any upper bound of i and j in I. Note the following.
    1. (\cdot is well defined) Let [(a_1,i_1)] = [(a_2,i_2)] and [(b_1,j_1)] = [(b_2,j_2)]. Then there exist r \geq i_1,i_2 and s \geq j_1,j_2 such that \rho_{i_1,r}(a_1) = \rho_{i_2,r}(a_2) and \rho_{j_1,s}(b_1) = \rho_{j_2,s}(b_2). Using the directedness of I, choose k_1 \geq i_1,j_1, k_2 \geq i_2,j_2, and t \geq k_1,k_2. Note the following.
      \rho_{k_1,t}(\rho_{i_1,k_1}(a_1) \cdot \rho_{j_1,k_1}(b_1))  =  \rho_{k_1,t}(\rho_{i_1,k_1}(a_1)) \cdot \rho_{k_1,t}(\rho_{j_1,k_1}(b_1))
       =  \rho_{i_1,t}(a_1) \cdot \rho_{j_1,t}(b_1)
       =  \rho_{r,t}(\rho_{i_1,r}(a_1)) \cdot \rho_{s,t}(\rho_{j_1,s}(b_1))
       =  \rho_{r,t}(\rho_{i_2,r}(a_2)) \cdot \rho_{s,t}(\rho_{j_2,s}(b_2))
       =  \rho_{i_2,t}(a_2) \cdot \rho_{j_2,t}(b_2)
       =  \rho_{k_2,t}(\rho_{i_2,k_2}(a_2)) \cdot \rho_{k_2,t}(\rho_{j_2,k_2}(b_2))
       =  \rho_{k_2,t}(\rho_{i_2,k_2}(a_2) \cdot \rho_{j_2,k_2}(b_2))

      Thus (\rho_{i_1,k_1}(a_1) \cdot \rho_{j_1,k_1}(b_1), t) \ \sigma\ (\rho_{i_2,k_2}(a_2) \cdot \rho_{j_2,k_2}(b_2),t), and in particular [(a_1,i_1)] \cdot [(b_1,j_1)] = [(a_2,i_2)] \cdot [(b_2,j_2)]. So \cdot is well-defined.

    2. (\cdot is associative) Let [(a,i)], [(b,j)], [(c,k)] \in \varinjlim A_i. Let r \geq i,j, s \geq j,k, and t \geq r,s. Then we have the following.
      \left( [(a,i)] \cdot [(b,j)] \right) \cdot [(c,k)]  =  [(\rho_{i,r}(a) \cdot \rho_{j,r}(b), r)] \cdot [(c,k)]
       =  [(\rho_{r,t}(\rho_{i,r}(a) \cdot \rho_{j,r}(b)) \cdot \rho_{k,t}(c), t)]
       =  [(\rho_{r,t}(\rho_{i,r}(a)) \cdot \rho_{r,t}(\rho_{j,r}(b)) \cdot \rho_{k,t}(c), t)]
       =  [(\rho_{i,t}(a) \cdot \rho_{j,t}(b) \cdot \rho_{k,t}(c), t)]
       =  [(\rho_{i,t}(a) \cdot \rho_{s,t}(\rho_{j,s}(b)) \cdot \rho_{s,t}(\rho_{k,s}(c)), t)]
       =  [(\rho_{i,t}(a) \cdot \rho_{s,t}(\rho_{j,s}(b) \cdot \rho_{k,s}(c)), t)]
       =  [(a,i)] \cdot [(\rho_{j,s}(b) \cdot \rho_{k,s}(c), s)]
       =  [(a,i)] \cdot \left( [(b,j)] \cdot [(c,k)] \right)

      Thus \cdot is associative.

    3. (\cdot distributes over +) We will show that \cdot distributes over + on the left; distributivity on the right is similar. Let [(a,i)], [(b,j)], [(c,k)] \in \varinjlim A_i, let r \geq j,k, and let t \geq i,r. Then we have the following.
      [(a,i)] \cdot \left( [(b,j)] + [(c,k)] \right)  =  [(a,i)] \cdot [(\rho_{j,r}(b) + \rho_{k,r}(c) ,r)]
       =  [(\rho_{i,t}(a) \cdot \rho_{r,t}(\rho_{j,r}(b) + \rho_{k,r}(c)), t)]
       =  [(\rho_{i,t}(a) \cdot \left( \rho_{r,t}(\rho_{j,r}(b)) + \rho_{r,t}(\rho_{k,r}(c)) \right), t)]
       =  [(\rho_{i,t}(a) \cdot \left( \rho_{j,t}(b) + \rho_{k,t}(c) \right), t)]
       =  [(\rho_{i,t}(a) \cdot \rho_{j,t}(b) + \rho_{i,t}(a) \cdot \rho_{k,t}(c), t)]
       =  [(\rho_{t,t}(\rho_{i,t}(a) \cdot \rho_{j,t}(b)) + \rho_{t,t}(\rho_{i,t}(a) \cdot \rho_{k,t}(c)), t)]
       =  [(\rho_{i,t}(a) \cdot \rho_{j,t}(b), t)] + [(\rho_{i,t}(a) \cdot \rho_{k,t}(c), t)]
       =  [(a,i)] \cdot [(b,j)] + [(a,i)] \cdot [(c,k)]

      Thus \cdot distributes over +.

    Thus (\varinjlim A_i, +, \cdot) is a ring. Moreover, we have the following.

    1. (If all the A_i are commutative, then \varinjlim A_i is commutative) Suppose the A_i are all commutative. Let [(a,i)], [(b,j)] \in \varinjlim A_i, and let k \geq i,j. Then we have the following.
      [(a,i)] \cdot [(b,j)]  =  [(\rho_{i,k}(a) \cdot \rho_{j,k}(b), k)]
       =  [(\rho_{j,k}(b) \cdot \rho_{i,k}(a), k)]
       =  [(b,j)] \cdot [(a,i)]

      Thus \varinjlim A_i is a commutative ring.

    2. (If all the A_i have 1 \neq 0 and the \rho_{i,j} are unital, then \varinjlim A_i has a multiplicative identity) Note first that – because the \rho_{i,j} are unital ring homomorphisms – \rho_{i,k}(1) = 1 = \rho_{j,k}(1) whenever k \geq i,j. Thus [(1,i)] = [(1,j)] for all i,j. Define 1 = [(1,i)]. Now let [(a,i)] \in \varinjlim A_i. Then 1 \cdot [(a,i)] = [(1,i)] \cdot [(a,i)] = [\rho_{i,i}(1) \cdot \rho_{i,i}(a), i)] = [(1 \cdot a,i)] = [(a,i)]. Similarly, [(a,i)] \cdot 1 = [(a,i)]. Thus 1 is a multiplicative identity in \varinjlim A_i.
    3. (If 1 \neq 0 for all A_1, then 1 \neq 0 in \varinjlim A_i) Suppose [(0,i)] = [(1,i)]. Then there exists j \geq i such that \rho_{i,j}(0) = \rho_{i,j}(1), so that 0 = 1 in A_j – a contradiction. Thus 1 \neq 0 in \varinjlim A_i.

    Thus if the A_i are commutative rings with 1 \neq 0, then \varinjlim A_i is a commutative ring with 1 \neq 0.

  5. Suppose now that C is an abelian group and that we have an indexed family of group homomorphisms \varphi_i : A_i \rightarrow C such that \varphi_i = \varphi_j \circ \rho_{i,j} for all i,j \in I.
    1. (Existence) Define \varphi : \varinjlim A_i \rightarrow C by \varphi([(a,i)]) = \varphi_i(a). We need to show that \varphi is a well defined group homomorphism.
      1. (Well-defined) Suppose [(a,i)] = [(b,j)]. Then there exists k \geq i,j such that \rho_{i,k}(a) = \rho_{j,k}(b). Since each latex \rho_{i,j}$ is well defined, \varphi_k(\rho_{i,k}(a)) = \varphi_k(\rho_{j,k}(b)). Then (\varphi_k \circ \rho_{i,k})(a) = (\varphi_k \circ \rho_{j,k})(b), and we have \varphi_i(a) = \varphi_j(b). Hence \varphi([(a,i)]) = \varphi([b,j)]), and \varphi is well defined.
      2. (Group Homomorphism) Let [(a,i)], [(b,j)] \in \varinjlim A_i, and let k \geq i,j. Then we have the following.
        \varphi([(a,i)] + [(b,i)])  =  \varphi([(\rho_{i,k}(a) + \rho_{j,k}(b), k)])
         =  \varphi_k(\rho_{i,k}(a) + \rho_{j,k}(b))
         =  \varphi_k(\rho_{i,k}(a)) + \varphi_k(\rho_{j,k}(b))
         =  (\varphi_k \circ \rho_{i,k})(a) + (\varphi_k \circ \rho_{j,k})(b)
         =  \varphi_i(a) + \varphi_j(b)
         =  \varphi([(a,i)]) + \varphi([(b,j)])

      So \varphi is a group homomorphism. Finally, note that for all i and all a \in A_i, (\varphi \circ \rho_i)(a) = \varphi(\rho_i(a)) = \varphi([(a,i)]) = \varphi_i(a). Thus \varphi \circ \rho_i = \varphi_i for all i.

    2. (Uniqueness) Suppose that we have a group homomorphism \psi : \varinjlim A_i \rightarrow C which also satisfies \psi \circ \rho_i = \varphi_i for all i \in I. Then for all i and all a \in A_i, we have \psi([(a,i)]) = \psi(\rho_i(a)) = (\psi \circ \rho_i)(a) = \varphi_i(a) = (\varphi \circ \rho_i)(a) = \varphi(\rho_i(a)) = \varphi([(a,i)]). Thus we have \psi = \varphi.

The set of all endomorphisms of an abelian group is a ring under pointwise addition and composition

Let A be any abelian group. Let R = \mathsf{Hom}(A,A) be the set of all group homomorphisms of the additive group A to itself, with addition defined pointwise and with multiplication defined as function composition. Prove that these operations make R into a ring with identity. Prove that the units of R are precisely the group automorphisms of A.


Let a,b,c \in R, and let x \in A.

  1. Note that (a+(b+c))(x) = a(x) + (b+c)(x) = a(x) + (b(x) + c(x)) = (a(x) + b(x)) + c(x) = (a+b)(x) + c(x) = ((a+b)+c)(x). Thus a+(b+c) = (a+b)+c, so that addition is associative.
  2. Consider the trivial homomorphism 0(x) = 0. If a is a homomorphism A \rightarrow A, then (0 + a)(x) = 0(x) + a(x) = a(x). Thus 0+a = a. Similarly, a + 0 = a, so that this set has an additive identity.
  3. Define the mapping \overline{a}(x) = -a(x). Since \overline{a}(x+y) = -a(x+y) = -a(x)-a(y) = \overline{a}(x) + \overline{a}(y), \overline{a} is a group homomorphism. Moreover, (a + \overline{a})(x) = a(x) + \overline{a}(x) = a(x) - a(x) = 0 = 0(x), so that a + \overline{a} = 0. Similarly, \overline{a} + a = 0. Thus every element has an additive inverse.
  4. Note that (a + b)(x) = a(x) + b(x) = b(x) + a(x) = (b+a)(x), so that a+b = b+a, and R is an abelian group under addition.
  5. Recall that function composition is always associative, so that multiplication is associative.
  6. Note that (a \circ (b+c))(x) = a((b+c)(x)) a(b(x) + c(x)) = a(b(x)) + a(c(x)) = (a \circ b)(x) + (a \circ c)(x) = (a \circ b + a \circ c)(x). Thus a \circ (b + c) = a \circ b + a \circ c, and multiplication distributes over addition on the left.
  7. Similarly, ((a + b) \circ c)(x) = (a+b)(c(x)) = a(c(x)) + b(c(x)) = (a \circ c)(x) + (b \circ c)(x) = (a \circ c + b \circ c)(x), and we have (a+b) \circ c = a \circ c + b \circ c. Thus multiplication distributes over addition on the right.
  8. Finally, let 1 denote the identity homomorphism. Clearly 1 \cdot a = a = a \cdot 1, so that 1 is a multiplicative identity element.
  9. If 1 = 0, then |A| = 1. If |A| = 1, then clearly 1 = 0. That is, 1 \neq 0 if and only if A is nontrivial.

Thus R is a ring with 1, and 1 \neq 0 if and only if A is nontrivial.

Now we consider the units in R.

If a \in R is a unit, then there exists a homomorphism b : A \rightarrow A such that a \circ b = b \circ a = 1. Thus a is injective and surjective, and in fact a is an automorphism of A. Conversely, if a is an automorphism of A, then a^{-1} is an automorphism, and we have a \circ a^{-1} = a^{-1} \circ a = 1. Thus a \in R is a unit. That is, the units of R are precisely the group automorphisms of A.

Characterize the natural numbers which are only the orders of abelian groups

Let n > 1 be an integer. Prove the following classification: every group of order n is abelian if and only if n = p_1^{a_1} \cdots p_k^{a_k}, p_i are distinct primes, a_i \in \{1,2\}, and p_i does not divide p_j^{a_j}-1 for all i and j.


(\Rightarrow) Suppose n has the properties above; we proceed by induction on k.

For the base case, let k = 1 and let G be a group of order n = p_1^{a_1}. Then G is either cyclic or elementary abelian, hence G is abelian.

For the inductive step, suppose that every group of width at most k which satisfies the given conditions for |G| is abelian. Let G be a group of width k+1 such that |G| satisfies the given properties. Now every proper subgroup of G has width at most k, so that by the induction hypothesis, every proper subgroup of G is abelian. By this previous exercise, G is solvable.

By this previous exercise, there is a (maximal) subgroup H \leq G such that H \leq G is normal and G/H has prime order; say G/H \cong Z_p. Let P be a Sylow p-subgroup of G. Now by FTFGAG, H is the direct product of its Sylow subgroups, and the Sylow subgroups of H are normal in H. In particular, H is generated by elements of prime and prime squared order. Let x \in H have prime or prime squared order and let y \in P. If \langle xy \rangle = G, then G is cyclic, hence abelian. If \langle xy \rangle is a proper subgroup of G, then xy = yx, so that \langle x \rangle \leq C_G(y). In either case, we have \langle x \rangle \leq C_G(P). Thus H \leq C_G(P), and we have HP = G \leq C_G(P). Thus P \leq Z(G). Similarly, HP = G \leq C_G(H), so that H \leq Z(G). Thus HP = G \leq Z(G), and G is abelian.

(\Leftarrow) Suppose every group of order n = p_1^{a_1} \cdots p_k^{a_k} is abelian. If (without loss of generality) a_1 \geq 3, let Q be a nonabelian group of order p_1^{a_1}. (We may take Q to be the direct product of a nonabelian group of order p^3 and a cyclic group.) Then Q \times Z_{p_2^{a_2} \cdots p_k^{a_k}} is nonabelian of order n. Thus a_i \in \{1,2\} for all i. Now suppose (again without loss of generality) that p_1 divides p_2^{a_2} - 1. If a_2 = 1, then we may construct a nonabelian group Q of order p_1p_2. Then Q \times Z_{p_3^{a_3} \cdots p_k^{a_k}} is a nonabelian group of order n. If a_2 = 2, we may construct a nonabelian group of order pq^2, following the patterns here or here (depending on whether p_1 does not or does divide p_2-1.) By taking the direct product with a cyclic group, there exists a nonabelian group of order n.

Thus, if every group of order n is abelian, then n has the properties enumerated in the problem statement.

For odd primes p, a p-group whose every subgroup is normal is abelian

Let p be an odd prime and let P be a p-group. Prove that if every subgroup of P is normal then P is abelian. (Note that Q_8 is a nonabelian 2-group with this property, so the result fails for p=2.)


[Disclaimer: I looked at Alfonso Gracia-Saz’s notes when solving this problem.]

We proceed by induction on k such that |P| = p^k.

For the base case, if k = 1, then P is cyclic and thus abelian.

For the inductive step, suppose the conclusion holds for all P such that |P| = p^k, where k \geq 1. Let P be a group of order p^{k+1}. If P is cyclic, then P is abelian. If P is not cyclic, then by the previous exercise there exists a normal subgroup U \leq P with U \cong Z_p \times Z_p. Choose x,y \in U such that U = \langle x \rangle \times \langle y \rangle. By the hypothesis, \langle x \rangle, \langle y \rangle \leq P are normal. Now P/\langle x \rangle and P/\langle y \rangle are groups of order p^k, and by the Lattice Isomorphism Theorem and the induction hypothesis, every subgroup of P/\langle x \rangle and P/\langle y \rangle is normal; thus P/\langle x \rangle and P/\langle y \rangle are abelian. By this previous exercise, P/(\langle x \rangle \cap \langle y \rangle) \cong P is abelian.

Properties of a semidirect product of an abelian group by Cyc(2) via the inversion homomorphism

Let H be any abelian group and let K = \langle x \rangle \cong Z_2. Define \varphi : K \rightarrow \mathsf{Aut}(H) by \varphi(x)(h) = h^{-1}. (Recall that this mapping is indeed an automorphism since H is abelian.) We may then construct the semidirect product G = H \rtimes_\varphi K and identify H and K as subgroups of G.

Prove that every element of G \setminus H has order 2. Prove that G is abelian if and only if h^2 = 1 for all h \in H.


First, let (h,k) \in G \setminus H; then k \neq 1, so that in fact (h,k) = (h,x).

Now (h,x)(h,x) = (h \cdot \varphi(x)(h), x^2) = (hh^{-1},1) = (1,1) = 1. Hence every element of G \setminus H has order 2.

We prove the second conclusion in two parts.

Suppose G is abelian. Then (h,1)(h,x) = (h,x)(h,1) for all h \in H. Multiplying, we see that (h \cdot \varphi(1)(h),x) = (h \cdot \varphi(x)(h), x), so that (h^2,x) = (1,x). Comparing entries we have h^2 = 1 for all h \in H.

Suppose now that h^2 = 1 for all h \in H, and let (h_1,k_1), (h_2,k_2) \in G.

  1. If k_1 = k_2 = 1, then (h_1,k_1), (h_2,k_2) \in H so that these elements commute.
  2. If (without loss of generality) k_1 = x and k_2 = 1, then (h_1,x)(h_2,1) = (h_1 h_2^{-1},x) = (h_2h_1, x) = (h_2,1)(h_1,x). Thus these elements commute.
  3. If k_1 = k_2 = x, then (h_1,x)(h_2,x) = (h_1h_2^{-1},1) = (h_2h_1^{-1},1) = (h_2,x)(h_1,x). So these elements commute.

Thus G is abelian.