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| + | =[[HW3_MA453Fall2008walther|HW3]], Chapter 4, Problem 9, [[MA453]], Fall 2008, [[user:walther|Prof. Walther]]= | ||
| + | ==Problem Statement== | ||
| + | ''Could somebody please state the problem?'' | ||
| + | |||
| + | ---- | ||
| + | ==Discussion== | ||
| + | |||
I do not understand how to tell what a generator of a subgroup is? I think that the subgroups of Z20 are (1,2,4,5,10,20), but that also might not be right. Anyways I could use a little explanation please. | I do not understand how to tell what a generator of a subgroup is? I think that the subgroups of Z20 are (1,2,4,5,10,20), but that also might not be right. Anyways I could use a little explanation please. | ||
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Check the back of the book. Theres a selected answer/hint section. It gives some good information about the problem. The subgroups are given by (1,2,4,5,10,20), which are the generators. So I think you are on the right track. Hope that helps. | Check the back of the book. Theres a selected answer/hint section. It gives some good information about the problem. The subgroups are given by (1,2,4,5,10,20), which are the generators. So I think you are on the right track. Hope that helps. | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | There is a corollary to the Fundamental Theorem of Cyclic Groups on page 79 of the textbook that is really useful for this problem. | ||
| + | Say you want to find all subgroups of <math>Z_n</math>. The corrolary states that, for each positive divisor k of n, the set <math>\langle n/k \rangle</math> is the unique subgroup of <math>Z_n</math> of order k. It also states that these subgroups are the only ones <math>Z_n</math> has. | ||
| + | Hence, to enumerate the subgroups, just find all the positive integer divisors of n (in this case 20), and use them to generate the subgroups. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | Confusion... | ||
| + | So, does it mean generator = subgroup? I mean,... like for the example above, 1,2,4,5,10,20 are the generators and <1>,<2>,<4>.... are the subgroup??? Correct me if I'm wrong... Thanks | ||
| + | |||
| + | --[[User:Mmohamad|Mmohamad]] 21:07, 21 September 2008 (UTC) | ||
| + | |||
| + | |||
| + | It does not mean that generator = subgroup. You get the generators from the group and you get the subgroups from the generators. Your notation is correct. 1,2,4,... are the generators and <1>,<2>,<4>, ... are the subgroups. For example, 1 is a generator and the subgroup of 1 is = <1> which is in fact = {1,2,3,4,5,6,7,8,9,...., 0} in this case. | ||
| + | |||
| + | -Ozgur | ||
| + | ---- | ||
| + | ---- | ||
| + | [[HW3_MA453Fall2008walther|Back to HW3]] | ||
| + | |||
| + | [[Main_Page_MA453Fall2008walther|Back to MA453 Fall 2008 Prof. Walther]] | ||
Latest revision as of 17:07, 22 October 2010
HW3, Chapter 4, Problem 9, MA453, Fall 2008, Prof. Walther
Problem Statement
Could somebody please state the problem?
Discussion
I do not understand how to tell what a generator of a subgroup is? I think that the subgroups of Z20 are (1,2,4,5,10,20), but that also might not be right. Anyways I could use a little explanation please.
Check the back of the book. Theres a selected answer/hint section. It gives some good information about the problem. The subgroups are given by (1,2,4,5,10,20), which are the generators. So I think you are on the right track. Hope that helps.
There is a corollary to the Fundamental Theorem of Cyclic Groups on page 79 of the textbook that is really useful for this problem. Say you want to find all subgroups of $ Z_n $. The corrolary states that, for each positive divisor k of n, the set $ \langle n/k \rangle $ is the unique subgroup of $ Z_n $ of order k. It also states that these subgroups are the only ones $ Z_n $ has. Hence, to enumerate the subgroups, just find all the positive integer divisors of n (in this case 20), and use them to generate the subgroups.
Confusion... So, does it mean generator = subgroup? I mean,... like for the example above, 1,2,4,5,10,20 are the generators and <1>,<2>,<4>.... are the subgroup??? Correct me if I'm wrong... Thanks
--Mmohamad 21:07, 21 September 2008 (UTC)
It does not mean that generator = subgroup. You get the generators from the group and you get the subgroups from the generators. Your notation is correct. 1,2,4,... are the generators and <1>,<2>,<4>, ... are the subgroups. For example, 1 is a generator and the subgroup of 1 is = <1> which is in fact = {1,2,3,4,5,6,7,8,9,...., 0} in this case.
-Ozgur
