International Journal of Pure and Applied Mathematics ————————————————————————– Volume 48 No. 4 2008, 473-476
A CHARACTERIZATION OF SUBGROUPS Soon-Mo Jung Section of Mathematics College of Science and Technology Hong-Ik University Chochiwon, 339-701, KOREA e-mail:
[email protected]
Abstract: We introduce a simple criterion for deciding whether a given subset of a group is a subgroup. AMS Subject Classification: 20K27, 20E07 Key Words: subgroup, group, criterion for subgroups
1. Introduction and Main Result It is well known that a subset H 6= ∅ of a group G is a subgroup of G if and only if ab−1 ∈ H
(1)
for all a, b ∈ H. Equivalently, we may say that H 6= ∅ is a subgroup of G if and only if a−1 b ∈ H
(2)
for all a, b ∈ H. The criterion (1) or (2) is a useful method for deciding whether a given nonempty subset of a group is a subgroup. However, we may sometimes encounter difficulties in gaining the inverse elements of arbitrarily given elements of some groups. In particular, it is cumbersome to calculate the inverse matrices. Received:
April 24, 2008
c 2008, Academic Publications Ltd.
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We will introduce a new criterion for deciding whether a given subset of a group is a subgroup. As we shall see in the following theorem, no inverse element appears in the process of our criterion, which may be considered as an advantage in comparison with the existing criterion (1) or (2). Theorem 1. Let H be a nonempty subset of a group G. Then, H is a subgroup of G if and only if, for each g ∈ G, the existence of h ∈ H with gh ∈ H implies g ∈ H. Proof. First, assume that H is a subgroup of G. If g ∈ G and h ∈ H with gh ∈ H, then there exists an h1 ∈ H with gh = h1 ∈ H, and hence g = h1 h−1 ∈ H. Conversely, assume that, for g ∈ G and h ∈ H, gh ∈ H implies g ∈ H. Let h1 , h2 ∈ H be arbitrary, and set g = h1 h−1 2 . Then, it is obvious that g = h1 h−1 ∈ G and gh = h ∈ H. Our hypothesis implies g = h1 h−1 2 1 2 2 ∈ H. In view of the criterion (1), we conclude that H is a subgroup of G. By a similar argument, but applying the criterion (2) instead of (1), we can prove the following theorem. Theorem 2. Let H be a nonempty subset of a group G. Then, H is a subgroup of G if and only if, for each g ∈ G, the existence of h ∈ H with hg ∈ H implies g ∈ H.
2. Examples We can find the following examples in usual textbooks for algebra, e.g., in [1–6], and we will apply our new criterion for subgroups to these examples for showing the utility of our criterion. Example 1. Let G be the group of nonzero real numbers under multiplication, and let H be the subset of positive rational numbers. Show that H is a subgroup of G. Solution. Let g and h be a nonzero real number and a positive rational number respectively such that gh is a positive rational number. In other words, assume that a g ∈ G and an h ∈ H satisfy gh ∈ H. Then, g must be a positive rational number, i.e., g ∈ H. According to Theorem 1, H is a subgroup of G.
A CHARACTERIZATION OF SUBGROUPS
Example 2. Let G be the group of all real 2 × 2 matrices
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a b c d
, with
ad − bc 6= 0 under matrix multiplication. Let a b H= ∈ G | ad 6= 0 . 0 d Verify that H is a subgroup of G. a b a1 b1 Solution. Let A = ∈ H. Since ∈ G and B = c d 0 d1 aa1 ab1 + bd1 AB = , ca1 cb1 + dd1 where ad − bc 6= 0 and a1 d1 6= 0, if we assume AB ∈ H, then c = 0 and hence A belongs to H. Due to our criterion, H is a subgroup of G. If we make use of the criterion (1) or (2) to solve this example, we should have made more effort because we should first calculate the inverse of B. Example 3. Let G be the group of all nonzero complex numbers a + bi (a, b real, not both 0) under multiplication, and let H = {a + bi ∈ G | a2 + b2 = 1}. Prove that H is a subgroup of G. Solution. Assume that a1 +b1 i (a1 , b1 real) is a nonzero complex number and that a2 +b2 i is an element of H. Furthermore, assume that (a1 +b1 i)(a2 +b2 i) ∈ H. Then, we get (a1 a2 − b1 b2 )2 + (a1 b2 + a2 b1 )2 = 1 and hence a21 + b21 = 1. This implies that a1 + b1 i also belongs to H. Now, by Theorem 1, H is a subgroup of G. Example 4. Let G be a group and a ∈ G. We define the normalizer N (a) of a in G by N (a) = {x ∈ G | xa = ax}. Verify that N (a) is a subgroup of G. Solution. Let g ∈ G and h ∈ N (a) be given. If gh belongs to N (a), then agh = gha. Since ha = ah, we have agh = gha = gah. Applying cancellation law to the last equation, we get ag = ga, and hence g ∈ N (a). According to Theorem 1, we conclude that N (a) is a subgroup of G.
References [1] R.A. Dean, Classical Abstract Algebra, Harper and Row, New York (1990).
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[2] J. Fraleigh, A First Course in Abstract Algebra, 4-th Edition, AddisonWesley, Reading, Mass (1989). [3] I.N. Herstein, Topics in Algebra, John Wiley and Sons, New York (1975). [4] T.W. Hungerford, Algebra, Springer, New York (1974). [5] S. Lang, Algebra, Addison-Wesley (1993). [6] G. Smith, O. Tabachnikova, Topics in Group Theory, Springer, London (2000).
