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Direct and Semidirect Products and Abelian Groups

Direct Products

Definition 1.  

1. The direct product $G_1 \times G_2 \times \cdots \times G_n$ of the groups $G_1, G_2, \ldots , G_n$ with operations $\star_1,\star_2, \ldots , \star_n$, respectively, is the set of $n$-tuples $(g_1,g_2, \ldots , g_n)$ where $g_i\in G_i$ with the operation defined componentwise:

   $$(g_1,g_2, \ldots ,g_n)\star (h_1,h_2, \ldots ,h_n) = (g_1 \star_1 h_1, g_2 \star_2 h_2. \ldots g_n\star_n h_n).$$

2. Similarly, the direct product $G_1 \times G_2 \times \cdots$ of the groups $G_1, G_2, \ldots$ with operations $\star_1,\star_2, \ldots$, respectively, is the set of sequences $(g_1,g_2, \ldots)$ where $g_i\in G_i$ with the operation defined componentwise:

   $$(g_1,g_2, \ldots)\star (h_1,h_2, \ldots) = (g_1 \star_1 h_1, g_2 \star_2 h_2. \ldots).$$

Proposition 1. If $G_1, \ldots, G_n$ are groups, their direct product is a group of order
$|G_1||G_2|\cdots |G_n|$ (if any $G_i$ is infinite, so is the direct product).

Proposition 2. Let $G_1, G_2, \ldots , G_n$ be group and let $G = G_1 \times G_2 \times \cdots \times G_n$ be their direct product.

1. For each fixed $i$ the set of elements of $G$ which have the identity of $G_j$ in the $j^{\text{th}}$ position for all $j \neq i$ and arbitrary elements of $G_i$ in position $i$ is a subgroup of $G$ isomorphic $G_i$:

   $$G_i \cong {(1,1,\ldots, 1, g_i,1,\ldots, 1) \mid g_i\in G_i},$$

   (here $g_i$ appears in the $i^{\text{th}}$ position). If we identity $G_i$ with this subgroup, then $G_i \trianglelefteq G$ and

   $$G/G_i \cong G_1\times \cdots \times G_{i-1} \times G_{i+1} \times \cdots \times G_n.$$

2. For each fixed $i$ define $\pi_i \colon G \to G_i$ by

3. $$\pi_i((g_1,g_2,\ldots,g_n)) = g_i.$$

   Then $\pi_i$ is a surjective homomorphism with

   $$\begin{aligned} \text{ker}\pi_i &= {(g_1,g_2, \ldots , g_{i-1}, 1, g_{i+1}) \mid g_j \in G_j \text{ for all } j\neq i} \ &\cong G_1\times \cdots \times G_{i-1} \times G_{i+1} \times \cdots \times G_n

   \end{aligned}$$

   (here 1 appears in position $i$).

4. Under the identifications in part 1, if $x \in G_i$ and $y\in G_j$ for some $i \neq j$, then $xy = yx$.

The Fundamental Theorem of Finitely Generated Abelian Groups

Definition 2.  

1. A group $G$ is finitely generated if there is some finite subset $A$ of $G$ such that $G = \langle A \rangle$.

2. For each $r\in \mathbb{Z}$ with $r \geq 0$ let $\mathbb{Z}^r = \mathbb{Z}\times \mathbb{Z}\times \cdots \times \mathbb{Z}$ be the direct product of r copies of the group $\mathbb{Z}$, where $\mathbb{Z}^0 = 1$. The group $\mathbb{Z}^r$ is called the free abelian group of order $r$.

Theorem 3. (The Fundamental Theorem of Finitely Generated Abelian Groups) Let $G$ be a finitely generated abelian group. Then

1. $$G \cong \mathbb{Z}^r \times Z_{n_1} \times Z_{n_2} \times \cdots \times Z_{n_s},$$

   for some $r,n_1,n_2, \ldots , n_s$ satisfying the following conditions:

   1. $r \geq 0$ and $n_j \geq 2$ for all $j$, and

   2. $n_{i+1} \mid n_i$ for all $1 \leq i \leq s-1$

2. the expression in 1. is unique: if $G \cong \mathbb{Z}^t \times Z_{m_1} \times Z_{m_2} \times \cdots \times Z_{m_u}$, where $t$ and $m_1, m_2, \ldots , m_u$ satisfy (a) and (b), then $t = r$ and $m_i = n_i$ for all $i$.

Definition 3. The integer $r$ in Theorem 3 is called the free rank or Betti number of $G$ and the integers $n_1, n_2, \ldots , n_s$ are called the invariant factors of $G$. The description of $G$ in Theorem 3(1) is called the invariant factor decomposition of $G$.

Note 1. There is a bijection between the set of isomorphism classes of finite abelian groups of order $n$ and the set of integer sequences $n_1, n_2, \ldots , n_s$ such that

1. $n_j \geq 2$ for all $j \in {1, 2, \ldots , s}$,

2. $n_{i+1} \mid n_i, 1 \leq i \leq s-1$, and

3. $n_1 n_2 \cdots n_s = n$.

Also notice that every prime divisor of $n$ must be a divisor of $n_1$ due to (2).

cor Corollary 4. If $n$ is the product of distinct primes, then up to isomorphism the only abelian group of order $n$ is the cyclic group of order $n$, $Z_n$.

Theorem 5. Let $G$ be an abelian group of order $n > 1$ and let the unique factorization of $n$ into distinct prime powers be

$$n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}.$$

Then

1. $G \cong A_1 \times A_2 \times \cdots \times A_k$, where $|A_i| = p_i^{\alpha_i}$

2. for each $A \in { A_1, A_2, \ldots , A_k }$ with $\vert A \vert = p^{\alpha}$ ,

   $$A \cong Z_{p^{\beta_1}} \times Z_{p^{\beta_2}} \times \cdots \times Z_{p^{\beta_t}}$$

   with $\beta_1 \geq \beta_2 \geq \ldots \geq \beta_t \geq 1$ and $\beta_1 + \beta_2 + \ldots + \beta_t = \alpha$ (where $t$ and $\beta_1, \beta_2, \ldots , \beta_t$ depend on $i$)

3. the decomposition in 1. and 2. is unique, i.e., if $G \cong B_1 \times B_2 \times \cdots \times B_m$, with $|B_i| = p_i^{\alpha_i}$ for all $i$, then $B_i \cong A_i$ and $B_i$ and $A_i$ have the same invariant factors.

Definition 4. The integers $p^{\beta_j}$ described in the proceeding theorem are called the elementary divisors of $G$. The description of $G$ in Theorem 5(1) and 5(2) is called the elementary divisor decomposition of $G$.

Note 2. For a group of order $p^\beta$ the invariant factors will be $p^{\beta_1}, p^{\beta_2}, \ldots , p^{\beta_t}$ such that

1. $\beta_j \geq 1$ for all $j \in {1,2, \ldots ,t}$,

2. $\beta_i \geq \beta_{i+1}$ for all $i$, and

3. $\beta_1 + \beta_2 + \ldots + \beta_t = \beta$

Proposition 6. Let $m,n \in \mathbb{Z}^+$.

1. $Z_m \times Z_n \cong Z_{mn}$ if and only if $(m,n) = 1$.

2. If $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$ then $Z_n \cong Z_{p_1^{\alpha_1}} \times Z_{p_2^{\alpha_2}} \times \cdots \times Z_{p_k^{\alpha_k}}$.

Note 3. The group $F_{20}$ of order 20 has generators and relations

$$\langle x,y \mid x^4 = y^5 = 1, xyx^{-1} = y^2 \rangle.$$

This group is called the Frobenius group of order 20 and can be viewed as the subgroup $F_{20} = \langle (2354), (12345) \rangle$ of $S_5$.

Recognizing Direct Products

Definition 5. Let $G$ be a group, let $x,y \in G$ and let $A, B$ be nonempty subsets of $G$.

1. Define $[x,y] = x^{-1}y^{-1}xy$, called the commutator of $x$ and $y$.

2. Define $[A,B] = \langle [a,b] \mid a \in A, b\in B \rangle$, the group generated by commutators of elements of $A$ and from $B$.

3. Define $G' = \langle [x,y] \mid x,y \in G \rangle$, the subgroup of $G$ generated by commutators of elements from $G$, called the commutator subgroup of $G$.

Proposition 7. Let $G$ be a group, let $x,y \in G$ and let $H \leq G$. Then

1. $xy = yx[x,y]$ (in particular, $xy = yx$ if and only if $[x,y] =1$).

2. $H \trianglelefteq G$ if and only if $[H,G] \leq H$.

3. $\sigma [x,y] = [ \sigma(x), \sigma(y)]$ for any automorphism $\sigma$ of $G$, $G'$char$G$ and $G/G'$ is abelian

4. $G/G'$ is the largest abelian quotient of $G$ in the sense that if $H \trianglelefteq G$ and $G/H$ is abelian, then $G' \leq H$. Conversely, if $G' \leq H$, then $H \trianglelefteq G$ and $G/H$ is abelian.

5. If $\varphi\colon G \to A$ is any homomorphism of $G$ into an abelian group $A$, then $\varphi$ factors through $G'$.

Proposition 8. Let $H$ and $K$ be subgroups of the group $G$. The number of distinct ways of writing each element of the set $HK$ in the form $hk$, for some $h \in H$ and $k \in K$ is $|H \cap K|$. In particular, if $H \cap K = 1$, then each element of $HK$ can be written uniquely as the product $hk$, for some $h \in H$ and $k \in K$.

Theorem 9. Suppose $G$ is a group with subgroups $H$ and $K$ such that

1. $H$ and $K$ are normal in $G$, and

2. $H \cap K = 1$.

Then $HK \cong H \times K$.

Note 4. The above conditions are simply the necessary conditions to ensure that the map

$$\begin{aligned} \varphi\colon & HK \to H \times K \ & hk \mapsto (h,k)

\end{aligned}$$

is well defined and an isomorphism.

Definition 6. If $G$ is a group and $H$ and $K$ are normal subgroups of $G$ with $H \cap K = 1$, we call $HK$ the internal direct product of $H$ and $K$. We shall (when emphasis is called for) call $H \times K$ the external direct product pf $H$ and $K$. (The distinction here is purely notational by Theorem 9).

Semidirect Products

Theorem 10. Let $H$ and $K$ be groups and let $\varphi$ be a homomorphism from $K$ into Aut$(H)$. Let $\cdot$ denote the (left) action of $K$ on $H$ determined by $\varphi$. Let $G$ be the set of order pairs $(h,k)$ with $h \in H$ and $k \in K$ and define the following multiplication on $G$:

$$(h_1, k_1) (h_2, k_2) = (h_1 k_1 \cdot h_2, k_1 k_2).$$

1. This multiplication makes $G$ into a group of order $ \vert G \vert = \vert H \vert \vert K \vert $ .

2. The sets ${(h,1) \mid h \in H }$ and ${(1,k) \mid k \in K}$ are subgroups of $G$ and the maps $h \mapsto (h,1)$ for $h \in H$ and $k \mapsto (1,k)$ for $k \in K$ are isomorphisms of these subgroups with the groups $H$ and $K$ respectively;

   $$H \cong {(h,1) \mid h \in H } \quad \text{and} \quad K \cong {(1,k) \mid k \in K}.$$

Identifying $H$ and $K$ with their isomorphic copies in $G$ described in 2. we have

3. $H \trianglelefteq G$

4. $H \cap K = 1$

5. for all $h \in H$ and $k \in K$, $khk^{-1} = k \cdot h = \varphi(k)(h)$

Definition 7. Let $H$ and $K$ be groups and let $\varphi$ be a homomorphism from $K$ into Aut$(H)$. The group described in Theorem 10 is called the semidirect product of $H$ and $K$ with respect to $\varphi$ and will be denoted by $H \rtimes_\varphi K$ (when there is no danger of confusion we shall simply write $H \rtimes K$).

Proposition 11. Let $H$ and $K$ be groups and let $\varphi\colon K \to$Aut$(H)$ be a homomorphism. Then the following are equivalent:

1. the identity (set) map between $H \rtimes K$ and $H \times K$ is a group homomorphism (hence and isomorphism)

2. $\varphi$ is the trivial homomorphism from $K$ into Aut$(H)$

3. $K \trianglelefteq H \rtimes k$.

Theorem 12. Suppose $G$ is a group with subgroups $H$ and $K$ such that

1. $H \trianglelefteq G$, and

2. $H \cap K = 1$.

Let $\varphi\colon K \to$Aut$(H)$ be the homomorphism defined by mapping $k \in K$ to the automorphism of left conjugation by $k$ on $H$. Then $HK \cong H \rtimes K$. In particular, if $G = HK$ with $H$ and $K$ satisfying 1. and 2., then $G$ is the semidirect product of $H$ and $K$.

Definition 8. Let $H$ be a subgroup of the group $G$. A subgroup $K$ of $G$ is called a complement for $H$ in $G$ if $G = HK$ and $H \cap K = 1$.

Note 5. With the above terminology, the criterion for recognizing a semidirect product is simply that there must exist a complement for some proper normal subgroup of $G$.