0.1 Contents

• 1 Group Theory Note
  • 1.1 Introduction
  • 1.2 Group Actions
    • 1.2.1 Orbit and Stabilizer
    • 1.2.2 Burnside's Lemma
      • 1.2.2.1 Group With $2$ Conjugacy Classes
  • 1.3 Abelian Group
    • 1.3.1 Free Abelian Group
    • 1.3.2 Divisible and Reduced Groups
      • 1.3.2.1 Exercises

1 Group Theory Note

1.1 Introduction

This is my personal note on group theory.

1.2 Group Actions

1.2.1 Orbit and Stabilizer

1.2.2 Burnside's Lemma

1.2.2.1 Group With $2$ Conjugacy Classes

Group $G$ with $2$ conjugacy classes, are particularly interesting. As only finite groups have $2$ conjugacy classes, well be $C_2$. And any group with $2$ conjugacy classes is simple. Thus, it may serve as a good example to infinite simple groups.

Theorem 1. If $G$ is a group with $2$ conjugacy classes, then $G$ is simple.

Proof: As $1$ itself is a conjugacy class, the other conjugacy class must be all the other elements.

Let $N$ be a non-trivial normal subgroup of $G$. Take $x \in N$ such that $x\neq 1$. Then for any $y\in G$, $y\neq 1$, there exists $g\in G$ such that $gxg^{-1} = y$. Thus, $N=G$.■

Theorem 2. If $G$ is a finite group with $2$ conjugacy classes, then $G$ is isomorphic to $C_2$.

Proof: As $1$ itself is a conjugacy class, the other conjugacy class must be all the other elements.

By Burnside's lemma, we have the following equation:

$$2 = \frac{1}{|G|}\sum_{g\in G}|C(g)| = 1 + \frac{1}{|G|}\sum_{g\neq 1, g \in G}|C(g)|.$$

Where $C(g) = \{x\in G \mid gxg^{-1} = x\}$, which is the elements that is fixed by $g$. As $g$ must fix $1$ and itself, thus, for any $g\neq 1$, $|C(g)| \ge 2$.

Therefore, we have the following inequality:

$$2= 1 + \frac{1}{|G|}\sum_{g\neq 1, g \in G}|C(g)| \ge 1 + \frac{2}{|G|}(|G|-1) = 2.$$

Which then gives us the equality:

$$|G| \le 2.$$

■

1.3 Abelian Group

In this section, we will only focus on Abelian groups. And we will use the additive notation for the group operation.

1.3.1 Free Abelian Group

Definition 1. An Abelian group $F$ is free Abelian if it is a direct sum of infinite cyclic groups. More precisely, there is a subset $X \subseteq F$ of elements of infinite order, called a basis of $F$, with $F = \sum_{x\in X}$; i.e., $F = \sum\mathbb{Z}$.

Theorem 3. There is an infinite p-primary group $\mathbb{Z}(p^{\infty})$ each of whose proper subgroups is finite and cyclic.

Proof: We define a group $G$ have generators:

$$X = \{x_0,x_1,\dots,x_i,\dots\}, i\in\mathbb{N}$$

with relations:

$$\{ px_0, x_0 - px_1, x_1 - px_2, \dots, x_i - px_{i+1}, \dots \}$$

We let $F$ be the free Abelian group generated by $X$. And let $R \le F$ be the subgroup generated by the relations. Let $a_i = x_i + R \in F/R = G$, as $\{x_0,x_1,\dots\}$ form a basis of $F$, $\{a_0,a_1,\dots\}$ form a basis of $G$. And as $pa_0 = 0$, $pa_1 = a_0$, $pa_2 = a_1$, $\dots$, for every $a_i$, we have $p^{i+1}a_i = 0$. So $G$ is a p-primary group.

To prove $G$ is infinite, we prove that $a_i$ are distinct. Assume $a_i = a_j$ for some $i< j$. Then by previous reasoning, we have $a_i = p^{j-i} a_j$, which implies $(p^{j-i}-1) a_j= 0$. And previous reasoning also shows that $p^{j+1}a_j = 0$. Thus, the order of $a_j$ must divides $\gcd(p^{j+1} , p^{j-i}-1) = 1$. Which implies $a_j = 0$, which is a contradiction.

Take $H$ to be a proper subgroup of $G$. By previous argument, for all $j< i$, $a_j$ is in the cyclic group generated by $a_i$. Thus, we have

$$G = \bigcup_{i=0}^{\infty} \langle a_i \rangle.$$

If $H$ is finite, then there must be $n$ such that $H \subseteq \langle a_n \rangle$. Then $H$ is a subgroup of finite and cyclic group, thus $H$ is finite and cyclic.

If $H$ is infinite, then there must be a sequence of elements in $H$, say $\{h_1, h_2, \dots\}$, such that $h_i \in \langle a_{n_i} \rangle$, and $h_i \notin \langle a_{n_i-1} \rangle$. Assume $h_i = ka_{n_i}$, then $\gcd(k,p)$ must be $1$, as otherwise $h_i$ will be in $\langle a_{n_i-1} \rangle$. Then $h_i$ also generate $\langle a_{n_i} \rangle$, which implies $a_{n_i} \in H$. And thus, $a_i \in H, \forall i$, which implies $H = G$.■

1.3.2 Divisible and Reduced Groups

Definition 2. A group $G$ is called divisible if for every $a \in G$ and every positive integer $n$, there exists an element $b \in G$ such that $nb = a$.

Example 1. The group $\mathbb{Z}(p^\infty)$ is divisible.

We define a group $\mathbb{Z}(p^\infty)$ have generators:

$$X = \{x_0,x_1,\dots,x_i,\dots\}, i\in\mathbb{N}$$

with relations:

$$\{ px_0, x_0 - px_1, x_1 - px_2, \dots, x_i - px_{i+1}, \dots \}$$

We let $F$ be the free Abelian group generated by $X$. And let $R \le F$ be the subgroup generated by the relations. Let $a_i = x_i + R \in F/R = G$, as $\{x_0,x_1,\dots\}$ form a basis of $F$, $\{a_0,a_1,\dots\}$ form a basis of $G$. Also, $pa_0 = 0$, $pa_1 = a_0$, $pa_2 = a_1$, $\dots$, therefore we have, for all $j< i$, $a_j$ is in the cyclic group generated by $a_i$. Thus, we have

$$G = \bigcup_{i=0}^{\infty} \langle a_i \rangle.$$

We take $n=p^r m$ such that $\gcd(p,m)=1$. Then for any element $b \in \mathbb{Z}(p^\infty)$, we have $n$ such that $b \in \langle a_n \rangle$. Assume $b = ka_n$, we also take $c = ka_{n+r}$. Then, by the relation of generators, we have $p^r c = b$. As $m$ is co-prime with $p$, by Bézout's identity, there exists $x,y$ such that $mx + p^{n+r+1}y = 1$. We take $d = xc$. Then

$$\begin{aligned} nd &= p^r m xc \\ & = p^r(1-p^{n+r+1}y)c \\ & = p^r c - p^{n+2r+1}y c \\ & = b - 0 \\ & = b. \end{aligned}$$

Which finish the proof that $\mathbb{Z}(p^\infty)$ is divisible.

Theorem 4 (Injective Property, Baer, 1940). Let $D$ be a divisible group and let $A$ be a subgroup of a group $B$. If $f: A \rightarrow D$ is a homomorphism, then $f$ can be extended to a homomorphism $\phi: B \rightarrow D$; that is, the following diagram commutes:

Proof: This theorem can be proved by Zorn's lemma, or an equivalent proof of well-ordering principle.

Consider $\mathcal{F}$ to be the set of all pairs $(C, g)$, where $C$ is a subgroup of $B$ containing $A$ and $g: C \rightarrow D$ is a homomorphism extending $f$. Note that $\mathcal{F}$ is non-empty as $(A,f) \in \mathcal{F}$.

We define a partial order on $\mathcal{F}$ by $(C_1, g_1) \le (C_2, g_2)$ if $C_1 \subseteq C_2$ and $g_2|_{C_1} = g_1$. Then for any chain $\{(C_i, g_i)\}_{i\in I}$ in $\mathcal{F}$, we can define $C = \bigcup_{i\in I} C_i$ and $g: C \rightarrow D$ by $g(c) = g_i(c)$ for $c\in C_i$. It is easy to check $g$ is well-defined and $g$ is a homomorphism. Thus, $(C,g)$ is an upper bound of the chain.

By Zorn's lemma, there exists a maximal element $(B', g')$ in $\mathcal{F}$. If $B' \neq B$, then there exists $b \in B \setminus B'$. And we discuss whether $B'\cap \langle b \rangle=0$.

If $B'\cap \langle b \rangle=0$, then we can extend $g'$ to $B' + \langle b \rangle$ by defining $g'(b) = 0$. Which contradicts the maximality of $B'$.

If $B'\cap \langle b \rangle \neq 0$, then we can take $k$ to be the smallest positive integer such that $kb \in B'$. Then as $D$ is divisible, we can find $d \in D$ such that $kd = kb$. Then we can extend $g'$ to $B' + \langle b \rangle$ by defining $g'(b) = d$. And it is easy to check that $g'$ is a homomorphism. Which contradicts the maximality of $B'$.■

Follows from the injective property, we have the following corollary:

Corollary 1. If a divisible group $D$ is a subgroup of a group $B$, then $D$ is a direct summand of $B$.

Proof: We consider the following diagram:

As $D$ is divisible, we can extend $1_D$ to $\varphi: B \rightarrow D$. By theorem in group extension, we have $B = D \oplus \ker \varphi$.■

Definition 3. If $G$ is a group, then $dG$ is the subgroup of $G$ generated by all divisible subgroups of $G$.

Note that $dG$ is invariant under all automorphism of $G$. As image of any divisible subgroup is also divisible.

Lemma 1. For any group $F$, $dG$ is the unique maximal divisible subgroup of $G$.

Proof: We first prove that $dG$ is divisible. For any $a \in dG$ and $n \in \mathbb{N}$, there exists $H$ which is a divisible subgroup of $G$ such that $a \in H$. As $H$ is divisible, there exists $b \in H$ such that $nb = a$. And as $H$ is a subgroup of $dG$, $b \in dG$.

The maximal is obvious, as if there exists a divisible subgroup $H$ such that $dG \subset H \subset G$, then $H \subset dG$, by the definition of $dG$.

It is unique as if there exists another maximal divisible subgroup $H$, then $\langle H, dG \rangle$ is also a divisible subgroup, then $\langle H, dG \rangle \subset dG$.■

Definition 4. A group $G$ is called reduced if $dG = 0$.

Theorem 5. For every group $G$, there exists a decomposition $G = dG \oplus R$, where $dG$ is divisible and $R$ is reduced.

Proof: Since $dG$ is divisible, by corollary of injective property, $dG$ is a direct summand of $G$. Thus, there exists $R\le G$ such that $G = dG \oplus R$.

Next, we prove that $R$ is reduced. If $R$ is not reduced, then there exists a non-zero element $r \in R$ such that for every $n \in \mathbb{N}$, $r$ is divisible by $n$. Then take an element $a \in dG$, then $a+r \in dG \oplus R$ is also divisible by $n$ for every $n \in \mathbb{N}$. Which contradicts the maximally of $dG$.■

Every abelian group is an extension of the torsion group $tG$ by a torsion free subgroup, but $tG$ need not to be a direct summand of $G$. However, $dG$ is a direct summand of $G$.

Definition 5. Given a group $G$ define $G[n] = \{x\in G \mid nx = 0\}$.

Lemma 2. If $G$ and $H$ are divisible p-primary group, then $G \cong H$ if and only if $G[p] \cong H[p]$.

Proof: The "only if" part is obvious. For the "if" part, we define $\phi: G[p] \rightarrow H[p]$, to be the isomorphism between $G[p]$ and $H[p]$. Then $\phi$ is also a homomorphism from $G[p]$ to $H$. By previous lemma, we can extend $\phi$ to a homomorphism $\Phi: G \rightarrow H$. And we prove that $\Phi$ is injective first.

For injectivity, we prove this by induction on $n \ge 1$, that if $p^n x = 0$, and $\Phi(x) = 0$, then $x = 0$. The base case $n=1$ is obvious, as now $x \in G[p]$, which means $0=\Phi(x) = \phi(x)$, and as $\phi$ is isomorphic, $x = 0$. Now we assume the statement is true for $n-1$, and we prove it for $n$. If $p^n x = 0$, then $p^{n-1} (px) = 0$, also $\Phi(x) = 0$ must implies $\Phi(px) = 0$, thus by the induction hypothesis, $px = 0$, which implies $x = 0$ by the base case.

As $\Phi$ is injective, $\Phi: G \rightarrow \text{Im}(\Phi)$ is an isomorphism. And we define $\phi': \text{Im}(\Phi) \rightarrow G$ as the inverse of $\Phi$. Also, note that the inverse of $\phi'$ extends the inverse of $\phi$, as it is easy to check $\phi'\mid H[p] = \phi^{-1}$. As $\phi'$ is a homomorphism from a subgroup of $H$ to $G$, by previous lemma, we can extend $\phi'$ to a homomorphism $\Phi': H \rightarrow G$. As $\phi'$ is a surjection by definition, $\Phi'$ is also a surjection. And also as $\Phi'$ extends $\phi^{-1}$, apply a similar reasoning as above, we have $\Phi'$ is injective. Thus, $\Phi'$ is an isomorphism.■

Theorem 6. Every divisible group $D$ is a direct sum of copies of $\mathbb{Q}$ and of copies of $\mathbb{Z}(p^\infty)$ for various primes $p$.

Proof: It is easy to check that $tD$, the torsion group of $D$ is divisible, so $tD$ is a direct summand of $D$. Thus, there exists $V \subset D$ such that $D= tD \oplus V$. As now $V$ is torsion free, and divisible, it is a vector space over $\mathbb{Q}$, and thus, $V$ is a direct sum of copies of $\mathbb{Q}$.

Now we prove that $tD$ is a direct sum of copies of $\mathbb{Z}(p^\infty)$, for various primes $p$. For prime $p$, let $B_p = \{ x \in tD \mid p^nx = 0 \text{ for some } n \in \mathbb{N} \}$. Then $B_p$ is a divisible p-primary group. It is easy to check that $tD = \bigoplus_{p} B_p$. So, we only focus on $B_p$.

It is easy to check that the group $B_p[p]$ is now a vector space over $\mathbb{F}_p$, let $r_p$ be the dimension of $B_p[p]$. By previous lemma, $B_p$ is isomorphic to direct sum of $r_p$ copies of $\mathbb{Z}(p^\infty)$.■

There is an analogy between theorems about free Abelian groups and divisible groups, that may be formalized as follows. Given a commutative diagram containing exact sequences, then its dual diagram is the diagram obtained by reversing all the arrows. For example, the dual diagram of $0 \rightarrow A \rightarrow B$ is $B^* \rightarrow A^* \rightarrow 0$, and this leads on the say that "subgroup" and "quotient" are dual concepts.

Theorem 7. Every group $G$ can be embedded in a divisible group $D$.

Proof: As every Abelian group can be written as $F/R$, where $F$ is a free Abelian group and $R$ is a subgroup of $F$. Now $F = \sum \mathbb{Z}$, so that $F \le \sum \mathbb{Q}$. Hence, $G = F/R = (\sum\mathbb{Z})/R \le (\sum\mathbb{Q})/R$. As $(\sum\mathbb{Q})/R$ is a quotient of a divisible group, it is divisible.■

Corollary 2. A group $G$ is divisible if and only if it is a direct summand of any group containing it.

Proof: By previous theorem, we can embed $G$ in a divisible group $D$. Then $G$ is a direct summand of $D$. And any direct summand of divisible group is divisible.■

1.3.2.1 Exercises

Exercise 1. If $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is an exact sequence and if $A$ and $C$ are reduced, then $B$ is reduced.

Proof: We name the maps in the exact sequence as follows:

As $i$ is injective, we now assume $A \le B$. Assume $B$ is not reduced, then $dB \neq 0$. Then, $j(dB)$ is divisible, as it is a quotient of divisible group. Because $C$ is reduced, $j(dB) = 0$, which implies, $dB \le \ker j = \text{Im} i = A$. As $A$ is reduced, $dB = 0$.■

Exercise 2.

• Prove that $\mathbb{Q}/\mathbb{Z} = \sum_p \mathbb{Z}(p^\infty)$.
• Prove that $\mathbb{Q}/\mathbb{Z}[n] \cong \mathbb{Z}_n$.

Proof: As $\mathbb{Q}/\mathbb{Z}$ is a quotient of divisible group, it is divisible. Also given $\frac{a}{b} \in \mathbb{Q}/\mathbb{Z}$, we have $b \frac{a}{b} = 0$. Thus, $t\mathbb{Q}/\mathbb{Z} = \mathbb{Q}/\mathbb{Z}$, and $\mathbb{Q}/\mathbb{Z}$ contains no summand of $\mathbb{Q}$.

It is easy to check that:

$$\mathbb{Q}/\mathbb{Z}[p] = \{ \frac{a}{b} \in \mathbb{Q}/\mathbb{Z} \mid \gcd(a,b) = 1, b = p \}.$$

Then it is a cyclic group of order $p$, which is isomorphic to $\mathbb{Z}(p^\infty)[p]$. Then by previous theorem and lemma, we have, $\mathbb{Q}/\mathbb{Z} = \sum_p \mathbb{Z}(p^\infty)$.

For the second part, we factorize $n = p_1^{r_1} \dots p_k^{r_k}$. Then we have:

$$\begin{aligned} \mathbb{Q}/\mathbb{Z}[n] &= (\sum_p \mathbb{Z}(p^\infty))[n] \\ &= \sum_p (\mathbb{Z}(p^\infty)[n]) \\ &= \sum_{i = 1}^{k} (\mathbb{Z}(p_i^\infty)[p_i^{r_i}]) \\ &= \sum_{i = 1}^{k} \mathbb{Z}_{p_i^{r_i}} \\ &= \mathbb{Z}_n. \end{aligned}$$

■

Exercise 3. Prove that a group $G$ is divisible if and only if $G$ has the injective property.

Proof: The "only if" part is given by previous theorem. For the "if" part, we prove that if a group $D$ contains $G$, $G$ is a direct summand of $D$. And then by previous corollary, $G$ is divisible.

By injective property, we have the following diagram, where $\varphi$ is an extension of $1_G$.

Then by previous theorem about group extension, $G$ is a direct summand of $D$.■

Exercise 4. A group $G$ is divisible if and only if $G = pG$ for all prime $p$.

Proof: If $G$ is divisible, then for any element $g \in G$, as $G$ is divisible, there exists $h \in G$ such that $ph = g$. Thus, we have $G \le pG$. And as $pG \le G$ obviously, we have $G = pG$.

Then, if $G = pG$ for all prime $p$, then for any $n\in \mathbb{N}$, we do induction on $n$. The base case $n=1$ is obvious, any element $g \in G$ is divisible by $1$. Now we assume that any element $g \in G$ is divisible by all natural numbers less than $n$, and we prove that $g$ is divisible by $n$. If $n$ is a prime, then as $G = nG$, there exists $h \in G$ such that $nh = g$. If $n$ is not a prime, then there exists $n_1 \neq 1$ and $n_2 \neq 1$ such that $n = n_1 n_2$. As $g$ is now divisible by $n_1$ by induction hypothesis, there exists $h_1 \in G$ such that $n_1 h_1 = g$. And as $h_1$ is now divisible by $n_2$ by induction hypothesis, there exists $h_2 \in G$ such that $n_2 h_2 = h_1$. Thus, $n h_2 = g$.■

Exercise 5. A $p$-primary group $G$ is divisible if and only if $G=pG$.

Proof: If $G$ is divisible, then for any element $g \in G$, as $G$ is divisible, there exists $h \in G$ such that $ph = g$. Thus, we have $G \le pG$. And as $pG \le G$ obviously, we have $G = pG$.

Then, if $G = pG$, then for any $n\in \mathbb{N}$, we do induction on $n$. The base case $n=1$ is obvious, any element $g \in G$ is divisible by $1$. Now we assume that any element $g \in G$ is divisible by all natural numbers less than $n$, and we prove that $g$ is divisible by $n$. If $n = p$, then as $G = pG$, there exists $h \in G$ such that $ph = g$. If $n$ is a prime that is not $p$, then we assume the order of $g$ is $p^r$, and by Bézout identity, there exists $x,y$ such that $p^rx + ny = 1$. Then we have $n (yg) = ny g = (1-p^rx)g = g$, thus $g$ is divisible by $n$. If $n$ is not a prime, then there exists $n_1 \neq 1$ and $n_2 \neq 1$ such that $n = n_1 n_2$. As $g$ is now divisible by $n_1$ by induction hypothesis, there exists $h_1 \in G$ such that $n_1 h_1 = g$. And as $h_1$ is now divisible by $n_2$ by induction hypothesis, there exists $h_2 \in G$ such that $n_2 h_2 = h_1$.■

Exercise 6. The following are equivalent for a group $G$:

1. $G$ is divisible.
2. Every non-zero quotient of $G$ is infinite.
3. $G$ has no maximal proper subgroup.

Proof: From 1 to 2: If $G$ is divisible, then for any non-zero quotient $H$ of $G$, $H$ is divisible. And by previous theorem, $H$ is a direct summand of $\mathbb{Q}$, and $\mathbb{Z}(p^\infty)$, thus $H$ is infinite.

From 2 to 3: We assume $G$ has $H$ as a maximal proper subgroup, and derive a contradiction. As $H$ is maximal, $G/H$ is simple. And by 2, $G/H$ is infinite. Take $x \in G/H$, if $x$ has finite order, then $\langle x \rangle$ is a proper subgroup of $G/H$, contradicts the simplicity of $G/H$. If $x$ has infinite order, then $\langle 2x \rangle$ is a proper subgroup of $G/H$, contradicts the simplicity of $G/H$.

From 3 to 1: We assume that $G$ has no maximal proper subgroup, and $G$ is now not divisible, and tries to derive a contradiction. As $G$ is not divisible, then there exists $p$ such that $G \neq pG$. And we assume $\varphi: G \rightarrow G/pG$ is the canonical map. Then $G/pG[p] = G/pG$, obviously. Thus, $G/pG$ is a vector space over $\mathbb{F}_p$, we select a basis of $G/pG$, namely $\{x_1, x_2, \dots\}$. Then, we define $\phi: G/pG \rightarrow \mathbb{F}_p$ by $\phi(m_1 x_2 + m_2 x_2 + \dots) = m_1$. Then $\phi$ is a surjection. And $\phi\varphi: G \rightarrow \mathbb{F}_p$ is a surjection. Thus, $\ker{\phi\varphi}$ is a maximal proper subgroup of $G$, which contradicts the assumption.■

Exercise 7. If $G$ and $H$ are divisible groups each of which is isomorphic to a subgroup of the other, then $G \cong H$. Is this true if we drop the adjective "divisible"?

Proof: Let $\varphi: G \rightarrow H$ be the embedding of $G$ to the subgroup of $H$ that is isomorphic to $G$. And let $\psi: H \rightarrow G$ be the embedding of $H$ to the subgroup of $G$ that is isomorphic to $H$. Then for any prime $p$, we can restrict $\varphi$ and $\psi$ to $G[p]$ and $H[p]$, and we have $\varphi[p]: G[p] \rightarrow H[p]$ and $\psi[p]: H[p] \rightarrow G[p]$. As $G[p]$ and $H[p]$ are now vector spaces over $\mathbb{F}_p$, and as we have the embedding as above, the dimension of $G[p]$ is less than or equal to the dimension of $H[p]$, and vice versa. Thus, the dimension of $G[p]$ is equal to the dimension of $H[p]$. And thus, $G[p]$ is isomorphic to $H[p]$. By previous lemma, $tG$ is isomorphic to $tH$.

For the torsion free part, we define $G' = G/tG$ and $H' = H/tH$. And $\varphi' : G' \rightarrow H'$ by $\phi'(g + tG) = \varphi(g) + tH$, and $\psi' : H' \rightarrow G'$ by $\psi'(h + tH) = \psi(h) + tG$. To prove $\varphi'$ is well-defined, we assume $g + tG = g' + tG$, then there exists $t \in tG$ such that $g = g' + t$, thus we have:

$$\begin{aligned} \varphi'(g+tG) - \varphi'(g'+ tG) &= \varphi(g) - \varphi(g') + tH \\ &= \varphi(g') + \varphi(t) - \varphi(g') + tH \\ &= \varphi(t) + tH \\ \end{aligned}$$

As $t\in tG$, $\varphi(t) \in tH$, thus $\varphi'(g+tG) = \varphi'(g'+ tG)$. In the same way, we can prove $\psi'$ is well-defined.

We then check that $\varphi'$ is injective. Assume $\varphi'(g + tG) = 0 + tH$, then $\varphi(g) \in tH$. As $\varphi$ is an isomorphism between $G$ and $\text{Img}(\varphi)$, the order of $g$ is same as the order of $\varphi(g)$ in $\text{Img}(\varphi)$, thus $g \in tG$. In the same way, we can prove $\psi'$ is injective.

Also, $G/tG$ and $H/tH$ are divisible, thus are vector spaces over $\mathbb{Q}$, and as both $\varphi'$ and $\psi'$ are injective, the dimension of $G/tG$ is equal to the dimension of $H/tH$.

Thus, $G$ is isomorphic to $H$.

If we drop the adjective "divisible", then the statement is not true. For instance, we take $G = \mathbb{Q}^{\omega} = \prod_{i\in \mathbb{N}} \mathbb{Q}$. And $H = Z \times G$. Then $G$ is obviously divisible, but $H$ is not.

We define $\varphi: G \rightarrow H$ by $\varphi((q_i)_{i\in \mathbb{N}}) = (0, (q_i)_{i\in \mathbb{N}})$. And $\psi: H \rightarrow G$ by $\psi((z, (q_i)_{i\in \mathbb{N}})) = (z, (q_{i-1})_{i\in \mathbb{N}\setminus\{0\}})$. Then both $\varphi$ and $\psi$ are injective, which are both embedding. But $G$ is not isomorphic to $H$, as one is divisible and the other is not.■

Exercise 8.

1. Prove that the following groups are all isomorphic:
   1. $\mathbb{R}/\mathbb{Z}$.
   2. The circle group $\mathbf{T}$.
   3. $\prod_p\mathbb{Z}(p^{\infty})$.
   4. $\mathbb{R}\oplus(\mathbb{Q}/\mathbb{Z})$.
   5. $\mathbb{C}^{\times}$.
2. Prove that $t(\mathbb{C}^{\times})\cong \mathbb{Q}/\mathbb{Z}$.

Proof: The isomorphism between $\mathbb{R}/\mathbb{Z}$ and the circle group is obvious.

By consider a real number as decimal representation, the cardinality of $\mathbb{R}/\mathbb{Z}$ is the same as $\prod_p\mathbb{Z}(p^{\infty})$. And it is also easy to verify that the torsion group of $\mathbb{R}/\mathbb{Z}$ and $\prod_p\mathbb{Z}(p^{\infty})$ is isomorphic to $\sum_p\mathbb{Z}(p^{\infty})$. Thus, $(\mathbb{R}/\mathbb{Z})/t(\mathbb{R}/\mathbb{Z})$ and $(\prod_p\mathbb{Z}(p^{\infty}))/t(\prod_p\mathbb{Z}(p^{\infty}))$ must have the same cardinality, by looking at the cardinal number division. And we could say $(\mathbb{R}/\mathbb{Z})/t(\mathbb{R}/\mathbb{Z})$ and $(\prod_p\mathbb{Z}(p^{\infty}))/t(\prod_p\mathbb{Z}(p^{\infty}))$ have the say dimension, again by cardinal number arithmetic. Thus, they are isomorphic.

As $\mathbb{Q}/\mathbb{Z} \cong \sum_p\mathbb{Z}(p^{\infty})$, $(\mathbb{R}\oplus(\mathbb{Q}/\mathbb{Z})) / t(\mathbb{R}\oplus(\mathbb{Q}/\mathbb{Z})) = \mathbb{R}$. Thus, we only need to verify that $\mathbb{R}$ is isomorphic to $(\mathbb{R}/\mathbb{Z})/(\mathbb{Q}/\mathbb{Z})$, which is equal to $\mathbb{R}/\mathbb{Q}$. As $\mathbb{R}$ is an infinite dimensional vector space over $\mathbb{Q}$, quotient by $\mathbb{Q}$ will minus one dimension, thus, the dimension of $\mathbb{R}/\mathbb{Q}$ is equal the dimension of $\mathbb{R}$, which verify that $\mathbb{R}$ is isomorphic to $\mathbb{R}/\mathbb{Q}$.

As $\mathbb{C}^{\times}$ is isomorphic to $\mathbb{R}_{+}^{\times}\oplus\mathbf{T}$, which $\mathbb{R}_{+}^{\times}$ is the multiplicative group of positive real numbers. And $\mathbb{R}_{+}^{\times}$ is isomorphic to $\mathbb{R}$, by the natural logarithm map. Thus, $\mathbb{C}^{\times}\cong \mathbb{R} \oplus \mathbf{T}$, and $t(\mathbb{C}^{\times}) = \mathbb{Q}/\mathbb{Z}$. Also $(\mathbb{C}^{\times})/t(\mathbb{C}^{\times}) \cong \mathbb{R} \oplus ((\mathbb{R}/\mathbb{Z})/(\mathbb{R}/\mathbb{Z}))$. By above discussion, we already now that $(\mathbb{R}/\mathbb{Z})/(\mathbb{R}/\mathbb{Z}) \cong \mathbb{R}$, thus, $(\mathbb{C}^{\times})/t(\mathbb{C}^{\times}) \cong \mathbb{R} \oplus \mathbb{R}$. It is obvious that $\mathbb{R} \oplus \mathbb{R}$ is then a vector space over $\mathbb{Q}$, and the dimension of it is equal to square of the dimension of $\mathbb{R}$. By the cardinal number arithmetic, square of infinite cardinal number is still the same as the original infinite cardinal number. Thus, we have $\mathbb{R} \oplus \mathbb{R} \cong \mathbb{R}$. Then by previous result, we have $\mathbb{R}/\mathbb{Z} \cong \mathbb{C}^{\times}$.■