(New page: |<math>Z_12\oplus Z_4\oplus Z_15</math>| = lcm(12,4,15) =60. Let G = <math>Z_12\oplus Z_4\oplus Z_15</math>, and H be a subgroup of G with |H|=9. By Lagrange's theorem, if H is a subgroup ...) |
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Let G = <math>Z_12\oplus Z_4\oplus Z_15</math>, and H be a subgroup of G with |H|=9. | Let G = <math>Z_12\oplus Z_4\oplus Z_15</math>, and H be a subgroup of G with |H|=9. | ||
By Lagrange's theorem, if H is a subgroup of G, then |H| divides |G|; however, 9 does not divide 60, therefore there is no such subgroup. | By Lagrange's theorem, if H is a subgroup of G, then |H| divides |G|; however, 9 does not divide 60, therefore there is no such subgroup. | ||
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| + | <math>\scriptstyle|Z_{12}\oplus Z_4\oplus Z_{15}|</math> is actually <math>\scriptstyle 12*4*15\,\,=\,\,720</math>. This is because <math>\scriptstyle Z_{12}\oplus Z_4\oplus Z_{15}</math> represents the set of all triples (tuples of length 3) of the form <math>\scriptstyle(a,b,c)</math>, where <math>\scriptstyle a\,\in Z_{12}</math>, <math>\scriptstyle b\,\in Z_4</math>, and <math>\scriptstyle c\,\in Z_{15}</math>. If an ''element'' of <math>\scriptstyle Z_{12}\oplus Z_4\oplus Z_{15}</math> has ''components'' of orders 12, 4, and 15 (for example <math>\scriptstyle(1,1,1)</math>), then the order of that ''element'' will be <math>\scriptstyle lcm(12,4,15)\,\,=\,\,60</math>. | ||
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| + | To find a subgroup of <math>\scriptstyle Z_{12}\oplus Z_4\oplus Z_{15}</math> of order 9, note that the element <math>\scriptstyle 4</math> has order 3 in <math>\scriptstyle Z_{12}</math> and order 1 in <math>\scriptstyle Z_4</math>. Also, the element <math>\scriptstyle 5</math> has order 3 in <math>\scriptstyle Z_{15}</math>. So <math>\scriptstyle\langle 4\rangle\oplus\langle 4\rangle\oplus\langle 5\rangle\,\,=\,\,\{0,4,8\}\oplus\{0\}\oplus\{0,5,10\}</math> is an order 9 subgroup of <math>\scriptstyle Z_{12}\oplus Z_4\oplus Z_{15}</math>, because <math>\scriptstyle|\{0,4,8\}\oplus\{0\}\oplus\{0,5,10\}|\,\,=\,\,3*1*3\,\,=\,\,9</math>. | ||
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| + | :--[[User:Narupley|Nick Rupley]] 23:09, 8 October 2008 (UTC) | ||
Latest revision as of 19:09, 8 October 2008
|$ Z_12\oplus Z_4\oplus Z_15 $| = lcm(12,4,15) =60. Let G = $ Z_12\oplus Z_4\oplus Z_15 $, and H be a subgroup of G with |H|=9. By Lagrange's theorem, if H is a subgroup of G, then |H| divides |G|; however, 9 does not divide 60, therefore there is no such subgroup.
$ \scriptstyle|Z_{12}\oplus Z_4\oplus Z_{15}| $ is actually $ \scriptstyle 12*4*15\,\,=\,\,720 $. This is because $ \scriptstyle Z_{12}\oplus Z_4\oplus Z_{15} $ represents the set of all triples (tuples of length 3) of the form $ \scriptstyle(a,b,c) $, where $ \scriptstyle a\,\in Z_{12} $, $ \scriptstyle b\,\in Z_4 $, and $ \scriptstyle c\,\in Z_{15} $. If an element of $ \scriptstyle Z_{12}\oplus Z_4\oplus Z_{15} $ has components of orders 12, 4, and 15 (for example $ \scriptstyle(1,1,1) $), then the order of that element will be $ \scriptstyle lcm(12,4,15)\,\,=\,\,60 $.
To find a subgroup of $ \scriptstyle Z_{12}\oplus Z_4\oplus Z_{15} $ of order 9, note that the element $ \scriptstyle 4 $ has order 3 in $ \scriptstyle Z_{12} $ and order 1 in $ \scriptstyle Z_4 $. Also, the element $ \scriptstyle 5 $ has order 3 in $ \scriptstyle Z_{15} $. So $ \scriptstyle\langle 4\rangle\oplus\langle 4\rangle\oplus\langle 5\rangle\,\,=\,\,\{0,4,8\}\oplus\{0\}\oplus\{0,5,10\} $ is an order 9 subgroup of $ \scriptstyle Z_{12}\oplus Z_4\oplus Z_{15} $, because $ \scriptstyle|\{0,4,8\}\oplus\{0\}\oplus\{0,5,10\}|\,\,=\,\,3*1*3\,\,=\,\,9 $.
- --Nick Rupley 23:09, 8 October 2008 (UTC)
