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Direct Products
Given two groups G and H, it is possible to construct a new group from the Cartesian product of G and H, G \times H.
Direct Products
Given two groups \(G\) and \(H\), it is possible to construct a new group from the Cartesian product of \(G\) and \(H\), \(G \times H\). Conversely, given a large group, it is sometimes possible to decompose the group; that is, a group is sometimes isomorphic to the direct product of two smaller groups. Rather than studying a large group \(G\), it is often easier to study the component groups of \(G\).
External Direct Products
If \((G,\cdot)\) and \((H, \circ)\) are groups, then we can make the Cartesian product of \(G\) and \(H\) into a new group. As a set, our group is just the ordered pairs \((g, h) \in G \times H\) where \(g \in G\) and \(h \in H\). We can define a binary operation on \(G \times H\) by \[\begin{aligned}\end{aligned}\] that is, we just multiply elements in the first coordinate as we do in \(G\) and elements in the second coordinate as we do in \(H\). We have specified the particular operations \(\cdot\) and \(\circ\) in each group here for the sake of clarity; we usually just write \((g_1, h_1)(g_2, h_2) = (g_1 g_2, h_1 h_2)\).
Example
Let \({\mathbb R}\) be the group of real numbers under addition. The Cartesian product of \({\mathbb R}\) with itself, \({\mathbb R} \times {\mathbb R} = {\mathbb R}^2\), is also a group, in which the group operation is just addition in each coordinate; that is, \((a, b) + (c, d) = (a + c, b + d)\). The identity is \((0,0)\) and the inverse of \((a, b)\) is \((-a, -b)\).
Example
Consider \[\begin{aligned}\end{aligned}\]. Although \({\mathbb Z}_2 \times {\mathbb Z}_2\) and \({\mathbb Z}_4\) both contain four elements, they are not isomorphic. Every element \((a,b)\) in \({\mathbb Z}_2 \times {\mathbb Z}_2\) other than the identity has order \(2\), since \((a,b) + (a,b) = (0,0)\); however, \({\mathbb Z}_4\) is cyclic.
The group \(G \times H\) is called the external direct product of \(G\) and \(H\). Notice that there is nothing special about the fact that we have used only two groups to build a new group. The direct product \[\begin{aligned}\end{aligned}\] of the groups \(G_1, G_2, \ldots, G_n\) is defined in exactly the same manner. If \(G = G_1 = G_2 = \cdots = G_n\), we often write \(G^n\) instead of \(G_1 \times G_2 \times \cdots \times G_n\).
Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.
Internal Direct Products
The external direct product of two groups builds a large group out of two smaller groups. We would like to be able to reverse this process and conveniently break down a group into its direct product components; that is, we would like to be able to say when a group is isomorphic to the direct product of two of its subgroups.
Let \(G\) be a group with subgroups \(H\) and \(K\) satisfying the following conditions.
\(G = HK = \{ hk : h \in H, k \in K \}\);
\(H \cap K = \{ e \}\);
\(hk = kh\) for all \(k \in K\) and \(h \in H\).
Example
The group \(U(8)\) is the internal direct product of \[\begin{aligned}\end{aligned}\].
Example
The dihedral group \(D_6\) is an internal direct product of its two subgroups \[\begin{aligned}\end{aligned}\]. It can easily be shown that \(K \cong S_3\); consequently, \(D_6 \cong {\mathbb Z}_2 \times S_3\).
Example
Not every group can be written as the internal direct product of two of its proper subgroups. If the group \(S_3\) were an internal direct product of its proper subgroups \(H\) and \(K\), then one of the subgroups, say \(H\), would have to have order \(3\). In this case \(H\) is the subgroup \(\{ (1), (123), (132) \}\). The subgroup \(K\) must have order \(2\), but no matter which subgroup we choose for \(K\), the condition that \(hk = kh\) will never be satisfied for \(h \in H\) and \(k \in K\).
Example
The group \({\mathbb Z}_6\) is an internal direct product isomorphic to \(\{ 0, 2, 4\} \times \{ 0, 3 \}\).
We can extend the definition of an internal direct product of \(G\) to a collection of subgroups \(H_1, H_2, \ldots, H_n\) of \(G\), by requiring that
\(G = H_1 H_2 \cdots H_n = \{ h_1 h_2 \cdots h_n : h_i \in H_i \}\);
\(H_i \cap \langle \cup_{j \neq i} H_j \rangle = \{ e \}\);
\(h_i h_j = h_j h_i\) for all \(h_i \in H_i\) and \(h_j \in H_j\).
Sage can quickly determine if two permutation groups are isomorphic, even though this should, in theory, be a very difficult computation.
Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.
Symbols used here
Multiply a_k for k = 1 up to n.
x belongs to A; every element of A is in B.
In either; in both; in A but not B.
i² = −1.
The two sides are different.
Naturals, integers, rationals, reals, complex numbers.
Marks the point where the statement has been established.
n divides a − b; a and b have the same remainder.
b is a multiple of a; the largest number dividing both.
A set with an operation; the do-nothing element; the element that undoes g.
Same structure; the group of cosets of a normal subgroup N.
The remainders 0…n−1 with clock arithmetic.
The set of morphisms; do g then f.
Questions people ask
What is a group, in plain words?
A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group — that is where the idea came from.
What is the difference between a ring and a field?
A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.
নিজের চেষ্টা করো
Parts of this page are adapted from Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Condensed and re-explained here; errors are ours.
আরও Abstract Algebra
GroupsSubgroups, cosets and Lagrange's theoremCyclic groups and permutation groupsHomomorphisms, normal subgroups and quotient groupsRings and fieldsGalois theory: why the quintic has no formula