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The form of subgroups A (n, <[bar.a]>) and B(n, <[bar.a]>) on additive group [Z.sub.m].

1. Introduction

Let G be additive Abelian group and H is a subgroup of G. For positive integer n, we defined

A(n, H) = {x [member of] G: nx [member of] H} and B(n, H) = {y [member of] G: y = nx [there exists] x [member of] H}

[Z.sub.m] is the additive group of residue classes modulo m under the operation [bar.a] + [bar.b] = [bar.a + b] for [bar.a], [bar.b] [member of] [Z.sub.m].

If a [member of] G, then <[bar.a]> = {ka: k [member of] Z) is the subgroup generated by a.

2. Preliminaries

Theorem 2.1 Let [bar.a], [bar.b] [member of] [Z.sub.m], [bar.a] = [bar.b] if and only if m|(a - b).

Theorem 2.2 If [bar.a] [member of] [Z.sub.m] and k [member of] Z, then k[bar.a] = [bar.ak].

Theorem 2.3 Let a, b, c [member of] Z. If a|bc and gcd(a, b) = 1, then a|c.

Theorem 2.4 Let a, b, d, q, r [member of] Z.

If gcd(a, b) = d such that a = qd and b = rd, then gcd(q, r) = 1.

Theorem 2.5 If gcd(a, b) = d, then ax + by = d for some x, y [member of] Z.

Theorem 2.6 gcd(a, b) x lcm(a, b) = [absolute value of ab].

3. Properties of A(n, H) and B(n, H) on additive Abelian group G

Theorem 3.1 B(n, H) [subset or equal to] H [subset or equal to] A(n, H).

Proof. If y [member of] B(n, H), then y = nx for some x [member of] H. But H is closed, so nx [member of] H and y [member of] H.

If x [member of] H, then nx [member of] H, and we get that x [member of] A(n, H). Thus B(n, H) [subset or equal to] H [subset or equal to] A(n, H).

Theorem 3.2 A(1, H) = B(1, H) = H.

Proof. If x [member of] A(1,H), then 1x [member of] H that is x [member of] H. Thus A(1,H) [subset or equal to] H but H [subset or equal to] A(1, H). So A(1, H) = H.

If x [member of] H, then x = 1x and x [member of] B(1, H). So H [subset or equal to] B(1, H) but B(1, H) [subset or equal to] H.

Theorem 3.3 A(n, H) and B(n, H) are the subgroups of an additive Abelian group G.

Proof. Let x, y [member of] A(n, H). We get that nx [member of] H and n y [member of] H. Since H is closed and commute, so nx - ny [member of] H such that n(x - y) = nx - ny [member of] H, thus x - y [member of] A(n, H). So A(n, H) is the subgroup of G. Let x, y [member of] B(n, H). We get that x = na and y = nb for some a, b [member of] H. So a - b [member of] H and x - y = na - nb = n(a - b), so x - y [member of] B(n, H). Thus B(n,H) is the subgroup of G.

Remark 3.1 It is easy to proved that A(n, G) = G and B(n, {0)) = {0).

Theorem 3.4 Let [H.sub.1] and [H.sub.2] be subgroup of G, then

1. If [H.sub.1] [subset or equal to] [H.sub.2], then A(n, [H.sub.0] [subset or equal to] A(n, [H.sub.2]) and B(n, [H.sub.0] [subset or equal to] B(n, [H.sub.2]).

2. A(n, [H.sub.1] [intersection] [H.sub.2]) = A(n, [H.sub.1] [intersection] A(n, [H.sub.2]) and B(n, [H.sub.1] [intersection] [H.sub.2]) [subset or equal to] B(n, [H.sub.1]) [intersection] B(n, [H.sub.2]).

Proof. 1. Assume that [H.sub.1] [subset or equal to] [H.sub.2] and let x [member of] A(n, [H.sub.1]). Then nx [member of] [H.sub.1] but [H.sub.1] [subset or equal to] [H.sub.2], so nx [member of] [H.sub.2], so x [member of] A(n, [H.sub.2]). Thus A(n, [H.sub.1]) [subset or equal to] A(n, [H.sub.2]). Let ye B(n, [H.sub.1]) we get y = nx for some x [member of] [H.sub.1] but [H.sub.1] [subset or equal to] [H.sub.2], so x [member of] [H.sub.2]. Thus y [member of] B(n, [H.sub.2]), and so B(n, [H.sub.1]) [subset or equal to] B(n, [H.sub.2]).

2. [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Thus A(n, [H.sub.1] [intersection] [H.sub.2]) = A(n, [H.sub.1]) [intersection] A(n, [H.sub.2]).

Let y [member of] B(n, [H.sub.1] [intersection] [H.sub.2]). Then y = nx for some x [member of] [H.sub.1] [intersection] [H.sub.2], therefore y = nx such that x [member of] [H.sub.1] and x [member of] [H.sub.2]. So y = nx, x [member of] [H.sub.1] and y = nx, x [member of] [H.sub.2] we get that y [member of] B(n, [H.sub.1]) and y [member of] B(n, [H.sub.2]). Thus y [member of] B(n, [H.sub.1]) [intersection] B(n, [H.sub.2]). Hence B(n, [H.sub.1] [intersection] [H.sub.2]) [subset or equal to] B(n, [H.sub.1]) [intersection] B(n, [H.sub.2]).

Remark 3.2 If [H.sub.1] [subset or equal to] [H.sub.2], then B(n, [H.sub.1] [intersection] [H.sub.2]) = B(n, [H.sub.1]) [intersection] B(n, [H.sub.2]).

4. Main Results

For [bar.a] [member of] [Z.sub.m], <[bar.a]> is the subgroup of [Z.sub.m] which generated by [bar.a].

Theorem 4.1 B(n, <[bar.a]>) = <[bar.na]>.

Proof. Let [bar.y] [member of] B(n, <[bar.a]>), then [bar.y] = [bar.nx] for some x [member of] <[bar.a]>. Since [bar.x] [member of] <[bar.a]>, we get [bar.x] = [bar.ka] for some k [member of] [Z.sup.+] [union] {0}. So [bar.y] = [bar.nx] = nk[bar.a] = kn[bar.a] = k[bar.na] [member of] <[bar.na]>, thus B(n, <[bar.a]>) [subset or equal to] <[bar.na]>. Let [bar.y] [member of] <[bar.na]>, then [bar.y] = k[bar.na] for some k [member of] [Z.sup.+] [union] {0}. We get [bar.y] = k[bar.na] = kn[bar.a] = n(k[bar.a]) such that k[bar.a] [member of] <[bar.a]>. So [bar.y] [member of] B(n,<[bar.a]>), and thus <[bar.na]> [subset or equal to] B(n, <[bar.a]>). Hence B(n, <[bar.a]>) = <[bar.na]>.

Corollary 4.1 B(n, [Z.sub.m]) = <[bar.n]>.

Proof. We known that [Z.sub.m] = <[bar.1]>. Thus B(n, [Z.sub.m]) = B(n, <[bar.1]>) = <[bar.n x 1]> = <[bar.n]>.

Theorem 4.2 If a|m, then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Proof. Assume that a|m and let gcd(n, a) = d. We get d|n and d|a, therefore n = d[x.sub.1] and a = d[x.sub.2] for some [x.sub.1], [x.sub.2] [member of] [Z.sup.+]. And we get that a/d = [x.sub.2] and gcd([x.sub.1], [x.sub.2]) = 1. Since a|m so m = a[x.sub.3] for some [x.sub.3] [member of] [Z.sup.+]. Now I want to show that A(n, <[bar.a]>) = <[[bar.x].sub.2]>.

Let [bar.x] [member of] A(n, <[bar.a]>). We get n[bar.x] [member of] <[bar.a]>, so n[bar.x] = k[bar.a] for some k [member of] [Z.sup.+] [union] {0) and then [bar.nx] = [bar.ka]. Therefore m|(nx - ka), thus nx - ka = mq for some q [member of] Z. So nx = ka + mq = ka+ a[x.sub.3]q = a(k+ [x.sub.3]q), and then d[x.sub.1]x = d[x.sub.2](k+ [x.sub.3]q) so [x.sub.1]x = [x.sub.2](k+ [x.sub.3]q), that is [x.sub.2]|[x.sub.1]x. Since gcd([x.sub.1], [x.sub.2]) = 1, thus [x.sub.2]|x that is [bar.x] [member of] < [[bar.x].sub.2]>. Hence A(n, <[bar.a]>) [subset or equal to] <[[bar.x].sub.2]]>.

Let [bar.x] [member of] <[[bar.x].sub.2]>. Then [bar.x] = k[[bar.x].sub.2] for some k [member of] [Z.sup.+] [union] {0}, so n[bar.x] = nk[[bar.x].sub.2] = d[x.sub.1]k[[bar.x].sub.2] and we get [bar.nx] = [bar.d[x.sub.1]]k[[bar.x].sub.2]. Therefore m|(nx - d[x.sub.1]k[x.sub.2]) that is nx - d[x.sub.1]k[x.sub.2] = mq for some q [member of] Z. And we get nx = d[x.sub.2] k[x.sub.1] + mq = a k [x.sub.1] + a[x.sub.3]q = a(k [x.sub.1] + [x.sub.3]q), this mean a | nx. Thus n[bar.x] = [bar.nx] [member of] <[bar.a]>. So [bar.x] [member of] A(n, <[bar.a]>), and we get that <[[bar.x].sub.2] > [subset or equal to] A(n, <[bar.a]>). Hence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Lemma 4.3 Let [bar.a] [member of] [Z.sub.m], if gcd(a, m) = d then <[bar.a]> = <[bar.d]>.

Proof. Assume that gcd(a,m) = d. We get d|a and d|m, therefore a = d[k.sub.1] for some [k.sub.1] [member of] [Z.sup.+] [union] {0} and m = d[k.sub.2] for some [k.sub.2] [member of] [Z.sup.+].

Let [bar.x] [member of] <[bar.a]>. Then [bar.x] = k[bar.a] for some k [member of] [Z.sup.+] [union] {0), thus [bar.x] = k[bar.a] and then m|(x ka). Therefore x - ka = mq for some q [member of] Z, then x = ka + mq = kd[k.sub.1] + d[k.sub.2]q = d(k[k.sub.1] + [k.sub.2]q). So d|x and we get x [member of] <[bar.d]>. Thus <[bar.a]> [subset or equal to] <[bar.d]>.

Since gcd(a, m) = d, we get that ax+ my = d for some x, y [member of] Z. Then ax - d = - my = m(-y) such that -y [member of] Z. Thus m|(ax - d), we get [bar.ax] = [bar.d] and so <[bar.xa]> = <[bar.ax]> = <[bar.d]>

Note that, it is easy to see that <[bar.xa]> [subset or equal to] <[bar.a]>. Thus <bar.d> [subset or equal to]<[bar.a]>. Hence <[bar.a]> = <[bar.d]>.

Theorem 4.3 [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] where gcd(a,m) = d.

Proof. Since gcd(a,m) = d, we get that d|m and <[bar.a]> = <[bar.d]>. Thus [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] because d | m.

We known that gcd(n,d) x lcm(n,d) = nd, or gcd(n, d) = nd/lcm(n,d). Therefore d/gcd(n,d) = d/[nd/lcm(n,d)] = d x [lcm(n, d)/nd] = [lcm(n, d/n)]

Thus [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

Corollary 4.3 [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. And if gcd(n, m) = 1, then A(n, {[bar.0])} = {[bar.0]}.

Proof. We known that {[bar.0]} = <[bar.0]>, and since gcd(0,m) = m.

Thus [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

And if gcd(n, m) = 1, then A(n, {[bar.0])} = <[bar.m]/1> = <[bar.m]> = <[bar.0]> = {[bar.0]).

Example In [Z.sub.20] = {[bar.0], [bar.1], [bar.2], [bar.3],...,[bar.19])

B(18, <[bar.15]>) = <[bar.18 x 15]> = <[bar.270]> = <[bar.10]>.

A(18, <[bar.15]>) = ?

Since gcd(15, 20) = 5 and gcd(18, 5) = 1.

Thus A(18, <[bar.15]>) = <[bar.5]/1> = <[bar.5]>.

References

[1] Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery (1960). An Introduction To The Theory of Numbers. candana. John Wiley & Sons, Inc.

[2] Marie J. Weiss and Roy Dubisch. (1962). Higher Algebra for the Undergraduate. Japan.Toppan Printing company, LTD.

[3] R. P. Sullivan. (2004). Msc Projects on Semigroup. School of Mathematics & Statistics The University of Western Australias.

[4] Thomas W. Hungerford. (1974). Algebra. New York. Springer Verlag.

[5] W. Keith Nicholson. (1999). Introduction to Abstract algebra second Edition. Cannada. John Wiley & Sons, Inc.

Worachead Sommanee

School of Mathematics and statistics, Faculty of Science and Technology, Chaing Mai Rajabhat University, Chiang Mai 50300, Thailand

E-mail: sommanee.w@hotmail.com
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Author:Sommanee, Worachead
Publication:Global Journal of Pure and Applied Mathematics
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Geographic Code:9THAI
Date:Aug 1, 2010
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