## 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.