This is the fifth post of the Introduction to Abstract Algebra series, and the last post about group theory. In this post, we will finalize our tool box by introducing automorphisms, direct products, and semi-direct products. Finally, we can start to harvest our efforts with group classification.

Direct Products

Definition 5.1: Direct Products

If $G_1, G_2, \cdots, G_n$ are groups, the direct products of the groups, denoted by $G_1 \times G_2 \times \cdots \times G_n$, is the direct products of the sets with following binary operation, $(g_1, g_2, \cdots, g_n) \circ (g_1', g_2', \cdots, g_n') = (g_1 \circ_1 g_1', g_2 \circ_2 g_2', \cdots, g_n \circ_n g_n')$.

Proposition 5.2: Direct Products

If $G_1 \times G_2 \times \cdots \times G_n$ is a direct product of $n$ groups,

1. For each $i \in [n]$, there exists a subjective homomorphism $\pi_i: G_1 \times G_2 \times \cdots \times G_n \to G_i$, where $\pi_i((g_1, \cdots, g_n)) = g_i$.
2. For each $i \in [n]$, there exists a injective homomorphism $\psi_i: G_i \to G_1 \times G_2 \times \cdots \times G_n$, where $\psi_i(g_i) = (e_1, \cdots, g_i, \cdots, e_n)$.
3. $G_1 \times G_2 \times \cdots \times G_n \cong G_\sigma(1) \times G_\sigma(2) \times \cdots \times G_\sigma(n)$ for $\sigma \in S_n$.

Proposition 5.3: Recognition Principle of Direct Products (Internal Direct Products)

If $G$ is a group, and $H, K$ are two normal subgroups such that $HK = G$ and $H \cap K = \{e\}$, then $G \cong H \times K$. We say $G$ is the internal direct product of $H$ and $K$.

Automorphisms

Definition 5.4: Automorphism For a group $G$, an automorphism is an isomorphism from $G$ to itself. $Aut(G)$ denotes the set of all automorphisms.

Proposition 5.5: The Conjugation Action and Automorphism

1. If $H \trianglelefteq G$, the action of $G$ acting on $H$ by conjugation is a homomorphism of $G$ to $Aut(H)$. The kernel of the action is $C_G(H)$, and thus $G/C_G(H) \cong \text{ a subgroup of } Aut(G)$.
2. If $H \subset G$, $N_G(H)/C_G(H) \cong \text{ a subgroup of } Aut(G)$.

Proposition 5.6: $Aut(\mathbb Z/n \mathbb Z) \cong (\mathbb Z/n \mathbb Z)^{\times}$

Classification of Finitely Generated Abelian Groups

First, we need to understand that mathematicians consider only the abstract structure of groups. So as long as two groups are isomorphic, we consider them as the “same”, no matter the actual elements or operations of the specific group. We’d like to classify groups into their “representatives” up to isomporphism. For example, groups of order 4 are either isomorphic to $C_4$ or $C_2 \times C_2$ (called the Klein four-group).

Definition 5.7: Finitely Generated Groups

A group $G$ is finitely generated if there exists $g_1, \cdots, g_n \in G$ for some finite $n$ such that $G = \langle g_1, \cdots, g_n \rangle$.

Theorem 5.8: Chinese Remainder If $n_1, n_2, \cdots, n_r$ are relatively prime integers, $G_{n_1} \times G_{n_2} \times \cdots \times G_{n_r} \cong G_{n_1 \times \cdots \times n_r}$.

Theorem 5.9: Fundamental Theorem of Abelian Groups

If $G$ is a finitely generated abelian group, then we have $G \cong \mathbb Z \times \cdots \times \mathbb Z \times \mathbb Z/n_1 \times \cdots \mathbb Z/n_k$ such that $n_1, \cdots, n_k \geq 2$ and $n_i+1 \mid n_i$ for $i \in [k-1]$. This decomposition is unique.

Equivalently, put $|G| = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$, then $G \cong A_1 \times \cdots \times A_r$ where $A_i = p_i^{a_i}$. Moreover, for each $A_i$, $A_i \cong \mathbb Z/p^{b_1} \times \cdots \times \mathbb Z/p^{b_m}$ where $b_1 \geq \cdots \geq b_m$. This decomposition is unique.

Example 5.1: Fundamental Theorem of Abelian Groups

Let $G = \mathbb Z/8 \times \mathbb Z/6 \times \mathbb Z/15$, then $G \cong \mathbb Z/120 \times \mathbb Z/6 \cong \mathbb Z/8 \times \mathbb Z/2 \times \mathbb Z/3 \times \mathbb Z/3 \times \mathbb Z/5$.

Proof. The theorem can be proven by induction on the size of a group. We also need the technique of “taking a group to a power”. The full proof can be found here.

Semi-Direct Products

Definition/Theorem 5.10: Semi-Direct Products

Let $H$ and $K$ be groups and let $\phi$ be a homomorphism from $K$ to $Aut(H)$. Define the group $G = \{(h,k)|h\in H, k \in K\}$, with the binary operation $(h_1, k_1) \circ (h_2, k_2) = (h_1 \phi(k_1)(h_2), k_1k_2)$. $G$ is the semi-direct product of $H$ and $K$, denoted as $G = H \rtimes K$. We have that

1. $G$ is a group, and $|G| = |H||K|$.
2. $H \cong \{(h,e_K) \in G: h \in H\}, K \cong \{(e_H,k) \in G: k \in K\}$.
3. $H \trianglelefteq G$.
4. $H \cap K = e$.
5. $khk^{-1} = \phi(k)(h)$ for all $h \in H, k \in K$.

Proof. The first claim can be shown by directly checking the associativity and inverse of the binary operation, using the definitions of homomorphisms. The second claim can be shown by defining isomorphisms. With the second claim, we can identify $H$ with $\{(h,e_K) \in G: h \in H\}$ and $K$ with $\{(e_H,k) \in G: k \in K\}$, or $h \in H$ with $(h, e)$ and $k \in K$ with $(e, k)$. The following claims are thus straightforward to show.

Proposition 5.11: Recognition of Semi-Direct Products

If $G$ is a group, and $H$ is a normal subgroup and $K$ is a subgroup such that $HK = G$ and $H \cap K = \{e\}$, then $G \cong H \rtimes_\phi K$, where $\phi: K \to Aut(H)$ such that $\phi(k)(h) = khk^{-1}$.

Examples of Group Classification

Example 5.2: Groups of Order $pq$, $p, q$ Primes

Assume WLOG that $p < q$. Let $P \in Syl_p(G), Q \in Syl_q(G)$ be two subgroups. By the Sylow’s theorems, $n_q \equiv 1 \mod q$ and $n_q \mid p$, but we assume $p < q$, so $n_q = 1$. We also have $n_p \equiv 1 \mod p$ and $n_p \mid q$.

If $p \nmid q-1$, $n_p = 1$ is the only possibility. Then $G \cong P \times Q$, a cyclic group.

Else if $p \mid q-1$, it is possible that $n_p = q$. Then $G \cong Q \rtimes_\phi P$, where $\phi: P \to Aut(Q) \cong \mathbb Z/q-1$. $P$ is a cyclic group, say $P = \langle y \rangle$, and $Aut(Q)$ is also cyclic and contains a unique subgroup of order $p$ (by classification of finitely generated abelian groups), say $<\gamma> \in Q$. $\phi_i(y) = \gamma^i$ for $i = 0, \cdots, p-1$. When $i=0$, the trivial homomorphism gives the direct product $G \cong P \times Q$. The other $p-1$ homomorphisms gives $p-1$ isomorphic non-abelian groups $G \cong Q \rtimes_{\phi_i} P$.

Example 5.3: Groups of Order 30

First, factor out $30 = 2 \times 3 \times 5$. By the Sylow’s theorems, $n_3 = 1 \text{ or } 10, n_5 = 1 \text{ or } 6$. Either $n_3 = 1$ or $n_5 = 1$, otherwise the number of elements exceeds 30, which means one of them is normal in $G$, so the product of a 3-group and a 5-group is a group of order 15, and from Example 5.2, we have that it is isomorphic to $\mathbb Z/15$. Let the group be $H \cong \mathbb Z/15$. Note that $H$ is normal in $G$ since it has index of 2. Let $K$ be a group of order 2. Since $G = HK, H \cap K = e$, we have $G \cong H \rtimes_\phi K$, hwere $\phi: K \to Aut(H) \cong (\mathbb Z/15)^\times \cong \mathbb Z/4 \times \mathbb Z/2$. Note that $K$ contains only the identity and an element of order 2. Let $H = \langle a \rangle \times \langle b \rangle \cong \mathbb Z/5 \mathbb Z/3$. There are only three elements in $Aut(H)$ of order 2:

• $\psi_1 \in Aut(H)$, where $\psi_1(a) = a^{-1}$, $\psi_1(b) = b$
• $\psi_2 \in Aut(H)$, where $\psi_2(a) = a, \psi_2(b) = b^{-1}$
• $\psi_3 \in Aut(H)$, where $\psi_3(a) = a^{-1}, \psi_2(b) = b^{-1}$

Corresponding to these three elements are three homomorphisms:

• $G \cong Q \rtimes_{\phi_1} P \cong \mathbb Z/5 \times D_6$
• $G \cong Q \rtimes_{\phi_2} P \cong \mathbb Z/3 \times D_{10}$
• $G \cong Q \rtimes_{\phi_3} P \cong D_{30}$

References

[1] Siegel, Kyler. MATH GU4041 Introduction to Modern Algebra I. Department of Mathematics, Columbia University. 2019.

[2] Dummit, David and Foote, Richard. Abstract Algebra. Third Edition. John Wiley and Sons, Inc. 2004.