- #1
wubie
Group Theory: Find the order of ...
Hello,
This is a question I have been given for homework. I would post this in homework section of the forum, but I need more than help to find a solution - I want to understand it. Explanation and understanding it is much more important to me than whether I finish the assignment or not and whether I get the mark for it or not. At any rate here is the question:
Now I will post relevant definitions that I have been given in lecture.
Order of a Group: The order of a group is the number of elements in G.
Order of an Element: If a is an element of a group, then the order of a is the smallest positive integer n such that an = u (identity element), if there is such an integer. Otherwise, if no positive power of a is u, then a has an infinite order.
Definitinon: If a is an element of a group G, then the subgroup consisting of all the powers of a (pos., neg., zero) is denoted by <a>. <a> is called the cyclic subgroup generated by a.
From the definitions, it is clear that the order of the subgroup <a> is equal to ord(a).
First I don't understand / am confused about the notation G = <a>.
I understand G to be a group. I also know that by definition <a> is a subgroup of all the power of a. I know that the subgroup <a> has an order 0f 21 - that is, it has 21 elements. I know that a group can be a subgroup of itself. Does that mean that G is the subgroup of itself? I am confused/confusing myself.
Now to address part i):
I know that <a> has 21 elements. I would think that would mean a21 = u (identity element).
Now if x is an element where x = a7, then by definition of the order of an element, ord(x) = 3 since x3 = a21 = u. Correct?
By the same reasoning, ord(y) = 7 since y7 = a21 = u.
I would then think that ord(xy) = 1. Since xy = a7*a3 = a21 = u. Is this correct?
Now to address part ii):
Following the same reasoning as part i), I know that <a> has n elements. I also know that n = pq. So <a> has pq elements. Therefore since an = u, then apq = u.
Now if x is an element where x = ap, then by definition of the order of an element, ord(x) = q since xq = (ap)q = apq = u.
Similarly, ord(y) = p since yp = (aq)p = aqp = u.
Again, ord(xy) = 1 since xy = ap*aq = apq = an = u.
So did I do this correctly? More importantly, did I reason it out correctly properly utilizing the definitions and concepts?
This is my second question:
Let G be a group, and let x be and element of G. Assume that x has infinite order. Prove that every non-zero power of x also has infinite order. That is, prove that, if i does not equal 0, then xx has infinite order. (Hint: This is a simple proof by contradiction, What happens if (xi)j = u for some positive integer j?)
I am unsure about how to do this question or even start. I will be back when I come up with something. If anyone cares to give me some help starting this one off I would appreciate it. Thankyou.
Any input/constructive advice/additions to the above questions and solutions are very welcome and appreciated. Thankyou.
Hello,
This is a question I have been given for homework. I would post this in homework section of the forum, but I need more than help to find a solution - I want to understand it. Explanation and understanding it is much more important to me than whether I finish the assignment or not and whether I get the mark for it or not. At any rate here is the question:
Let G = <a> be cyclic or order 21. Consider the elements x = a7 and y = a3.
i) Find the orders of x, y, and xy.
ii) Generalize part i): Let p and q be different prime numbers, put n = pq, and let G = <a> be cyclic of order n. Put x = ap and y = aq. Find the orders of x, y, and xy.
Now I will post relevant definitions that I have been given in lecture.
Order of a Group: The order of a group is the number of elements in G.
Order of an Element: If a is an element of a group, then the order of a is the smallest positive integer n such that an = u (identity element), if there is such an integer. Otherwise, if no positive power of a is u, then a has an infinite order.
Definitinon: If a is an element of a group G, then the subgroup consisting of all the powers of a (pos., neg., zero) is denoted by <a>. <a> is called the cyclic subgroup generated by a.
From the definitions, it is clear that the order of the subgroup <a> is equal to ord(a).
First I don't understand / am confused about the notation G = <a>.
I understand G to be a group. I also know that by definition <a> is a subgroup of all the power of a. I know that the subgroup <a> has an order 0f 21 - that is, it has 21 elements. I know that a group can be a subgroup of itself. Does that mean that G is the subgroup of itself? I am confused/confusing myself.
Now to address part i):
I know that <a> has 21 elements. I would think that would mean a21 = u (identity element).
Now if x is an element where x = a7, then by definition of the order of an element, ord(x) = 3 since x3 = a21 = u. Correct?
By the same reasoning, ord(y) = 7 since y7 = a21 = u.
I would then think that ord(xy) = 1. Since xy = a7*a3 = a21 = u. Is this correct?
Now to address part ii):
Following the same reasoning as part i), I know that <a> has n elements. I also know that n = pq. So <a> has pq elements. Therefore since an = u, then apq = u.
Now if x is an element where x = ap, then by definition of the order of an element, ord(x) = q since xq = (ap)q = apq = u.
Similarly, ord(y) = p since yp = (aq)p = aqp = u.
Again, ord(xy) = 1 since xy = ap*aq = apq = an = u.
So did I do this correctly? More importantly, did I reason it out correctly properly utilizing the definitions and concepts?
This is my second question:
Let G be a group, and let x be and element of G. Assume that x has infinite order. Prove that every non-zero power of x also has infinite order. That is, prove that, if i does not equal 0, then xx has infinite order. (Hint: This is a simple proof by contradiction, What happens if (xi)j = u for some positive integer j?)
I am unsure about how to do this question or even start. I will be back when I come up with something. If anyone cares to give me some help starting this one off I would appreciate it. Thankyou.
Any input/constructive advice/additions to the above questions and solutions are very welcome and appreciated. Thankyou.
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