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Abstract Algebra: Theory and Applications

Thomas W. Judson

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Section 3.3 Subgroups

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Subsection Definitions and Examples

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Sometimes we wish to investigate smaller groups sitting inside a larger group. The set of even integers 2Z={…,−2,0,2,4,…} is a group under the operation of addition. This smaller group sits naturally inside of the group of integers under addition. We define a subgroup H of a group G to be a subset H of G such that when the group operation of G is restricted to H, H is a group in its own right. Observe that every group G with at least two elements will always have at least two subgroups, the subgroup consisting of the identity element alone and the entire group itself. The subgroup H={e} of a group G is called the trivial subgroup. A subgroup that is a proper subset of G is called a proper subgroup. In many of the examples that we have investigated up to this point, there exist other subgroups besides the trivial and improper subgroups.
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Example 3.24.

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Consider the set of nonzero real numbers, R∗, with the group operation of multiplication. The identity of this group is 1 and the inverse of any element a∈R∗ is just 1/a. We will show that
Q∗={p/q:pandqare nonzero integers}
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is a subgroup of R∗. The identity of R∗ is 1; however, 1=1/1 is the quotient of two nonzero integers. Hence, the identity of R∗ is in Q∗. Given two elements in Q∗, say p/q and r/s, their product pr/qs is also in Q∗. The inverse of any element p/q∈Q∗ is again in Q∗ since (p/q)−1=q/p. Since multiplication in R∗ is associative, multiplication in Q∗ is associative.
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Example 3.25.

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Recall that C∗ is the multiplicative group of nonzero complex numbers. Let H={1,−1,i,−i}. Then H is a subgroup of C∗. It is quite easy to verify that H is a group under multiplication and that H⊂C∗.
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Example 3.26.

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Let SL2(R) be the subset of GL2(R) consisting of matrices of determinant one; that is, a matrix
A=(abcd)
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is in SL2(R) exactly when ad−bc=1. To show that SL2(R) is a subgroup of the general linear group, we must show that it is a group under matrix multiplication. The 2×2 identity matrix is in SL2(R), as is the inverse of the matrix A:
A−1=(d−b−ca).
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It remains to show that multiplication is closed; that is, that the product of two matrices of determinant one also has determinant one. We will leave this task as an exercise. The group SL2(R) is called the special linear group.
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Example 3.27.

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It is important to realize that a subset H of a group G can be a group without being a subgroup of G. For H to be a subgroup of G, it must inherit the binary operation of G. The set of all 2×2 matrices, M2(R), forms a group under the operation of addition. The 2×2 general linear group is a subset of M2(R) and is a group under matrix multiplication, but it is not a subgroup of M2(R). If we add two invertible matrices, we do not necessarily obtain another invertible matrix. Observe that
(1001)+(−100−1)=(0000),
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but the zero matrix is not in GL2(R).
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Example 3.28.

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One way of telling whether or not two groups are the same is by examining their subgroups. Other than the trivial subgroup and the group itself, the group Z4 has a single subgroup consisting of the elements 0 and 2. From the group Z2, we can form another group of four elements as follows. As a set this group is Z2×Z2. We perform the group operation coordinatewise; that is, (a,b)+(c,d)=(a+c,b+d). Figure 3.29 is an addition table for Z2×Z2. Since there are three nontrivial proper subgroups of Z2×Z2, H1={(0,0),(0,1)}, H2={(0,0),(1,0)}, and H3={(0,0),(1,1)}, Z4 and Z2×Z2 must be different groups.
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+(0,0)(0,1)(1,0)(1,1)(0,0)(0,0)(0,1)(1,0)(1,1)(0,1)(0,1)(0,0)(1,1)(1,0)(1,0)(1,0)(1,1)(0,0)(0,1)(1,1)(1,1)(1,0)(0,1)(0,0)
Figure 3.29. Addition table for Z2×Z2
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Subsection Some Subgroup Theorems

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Let us examine some criteria for determining exactly when a subset of a group is a subgroup.

Proof.

First suppose that H is a subgroup of G. We must show that the three conditions hold. Since H is a group, it must have an identity eH. We must show that eH=e, where e is the identity of G. We know that eHeH=eH and that eeH=eHe=eH; hence, eeH=eHeH. By right-hand cancellation, e=eH. The second condition holds since a subgroup H is a group. To prove the third condition, let h∈H. Since H is a group, there is an element h′∈H such that hh′=h′h=e. By the uniqueness of the inverse in G, h′=h−1.
Conversely, if the three conditions hold, we must show that H is a group under the same operation as G; however, these conditions plus the associativity of the binary operation are exactly the axioms stated in the definition of a group.

Proof.

First assume that H is a subgroup of G. We wish to show that gh−1∈H whenever g and h are in H. Since h is in H, its inverse h−1 must also be in H. Because of the closure of the group operation, gh−1∈H.
Conversely, suppose that H⊂G such that H≠∅ and gh−1∈H whenever g,h∈H. If g∈H, then gg−1=e is in H. If g∈H, then eg−1=g−1 is also in H. Now let h1,h2∈H. We must show that their product is also in H. However, h1(h2−1)−1=h1h2∈H. Hence, H is a subgroup of G.
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