|$ 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)
