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 January 18th, 2013, 05:37 PM #1 Member   Joined: May 2012 Posts: 56 Thanks: 0 A question groups and subgroups... In this link: http://www.millersville.edu/~bikenaga/c ... /revs2.pdf For example 2, do I just write down several multiples of 6 and 15 and look at what those multiples are mod 8...or is there a more efficient way to do this? Also, for example 10, why does the fact that the largest possible order is 40, tell us that not element can generate the group $Z_8 x Z_{10}$? Thanks in advance
 January 19th, 2013, 11:48 PM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: A question groups and subgroups... when all else fails (that is when you can't think of a clever short-cut), one simply computes powers of the generator. since this group is additive (the operation is a kind of addition), we write "powers" as "multiples". so the "square" of (6,10) is: (6,10) + (6,10) = (4,5) (since 12 = 4 (mod 8), and 20 = 5 (mod 15)). the "cube" is thus: (6,10) + (6,10) + (6,10) = (4,5) + (6,10) = (2,0). to compute the 4th power, etc., we just keep adding (6,10) to our previous answer, and reduce the first coordinate mod 8, and the second coordinate mod 15. the example has the powers listed in order. for example 10: if a group has a single generator, it is cyclic, and this generator has order the order of G. that is: G cyclic <=> there exists x in G with G = <=> there exists x in G with |x| = |G|. so if Z8 x Z10 had a single generator (that is, if it is cyclic), this generator would have to have order 80, since Z8 x Z10 has 80 elements. but no such element exists, since the maximum order is 40 at most. (it turns out Z8 x Z10 does indeed have an element of order 40, as well: (1,2) is such an element: the divisors of 80 are 1,2,4,5,8,10,16,20,40 and 80 <---only possible orders, by lagrange's theorem. (1,2) is not the identity (not of order 1) (1,2) + (1,2) = (2,4) which is not the identity, so not of order 2. 4(1,2) = (2,4) + (2,4) = (4,8), so (1,2) is not of order 4. 5(1,2) = (4,8) + (1,2) = (5,0), so (1,2) is not of order 5. 8(1,2) = (4,8) + (4,8) = (0,6), so (1,2) is not of order 8. 10(1,2) = (5,0) + (5,0) = (2,0), so (1,2) is not of order 10. 16(1,2) = (0,6) + (0,6) = (0,2), so (1,2) is not of order 16. 20(1,2) = (2,0) + (2,0) = (4,0), so (1,2) is not of order 20. 40(1,2) = (4,0) + (4,0) = (0,0), thus (1,2) is of order 40 <--note the short-cuts i took to compute the multiples of (1,2). these work well when the power (multiple) we are raising to is composite, or close to a composite number. (for example for an odd prime p, we know that (p-1)/2 is an integer, and if we've already computed a^((p-1)/2) then a^p = [(a^((p-1)/2))^2](a). these are some of the "short-cuts" i hinted at in the beginning of this post).
January 21st, 2013, 07:27 AM   #3
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Re: A question groups and subgroups...

Quote:
 Originally Posted by Deveno when all else fails (that is when you can't think of a clever short-cut), one simply computes powers of the generator. since this group is additive (the operation is a kind of addition), we write "powers" as "multiples". so the "square" of (6,10) is: (6,10) + (6,10) = (4,5) (since 12 = 4 (mod , and 20 = 5 (mod 15)). the "cube" is thus: (6,10) + (6,10) + (6,10) = (4,5) + (6,10) = (2,0). to compute the 4th power, etc., we just keep adding (6,10) to our previous answer, and reduce the first coordinate mod 8, and the second coordinate mod 15. the example has the powers listed in order. for example 10: if a group has a single generator, it is cyclic, and this generator has order the order of G. that is: G cyclic <=> there exists x in G with G = <=> there exists x in G with |x| = |G|. Thank you so much for clarifying. so if Z8 x Z10 had a single generator (that is, if it is cyclic), this generator would have to have order 80, since Z8 x Z10 has 80 elements. but no such element exists, since the maximum order is 40 at most. (it turns out Z8 x Z10 does indeed have an element of order 40, as well: (1,2) is such an element: the divisors of 80 are 1,2,4,5,8,10,16,20,40 and 80 <---only possible orders, by lagrange's theorem. (1,2) is not the identity (not of order 1) (1,2) + (1,2) = (2,4) which is not the identity, so not of order 2. 4(1,2) = (2,4) + (2,4) = (4,, so (1,2) is not of order 4. 5(1,2) = (4, + (1,2) = (5,0), so (1,2) is not of order 5. 8(1,2) = (4, + (4, = (0,6), so (1,2) is not of order 8. 10(1,2) = (5,0) + (5,0) = (2,0), so (1,2) is not of order 10. 16(1,2) = (0,6) + (0,6) = (0,2), so (1,2) is not of order 16. 20(1,2) = (2,0) + (2,0) = (4,0), so (1,2) is not of order 20. 40(1,2) = (4,0) + (4,0) = (0,0), thus (1,2) is of order 40 <--note the short-cuts i took to compute the multiples of (1,2). these work well when the power (multiple) we are raising to is composite, or close to a composite number. (for example for an odd prime p, we know that (p-1)/2 is an integer, and if we've already computed a^((p-1)/2) then a^p = [(a^((p-1)/2))^2](a). these are some of the "short-cuts" i hinted at in the beginning of this post).

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