- #1

**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:

Let G = <a> be cyclic or order 21. Consider the elements x = a^{7}and y = a^{3}.

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 = a^{p}and y = a^{q}. 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 a

^{n}= 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 a

^{21}= u (identity element).

Now if x is an element where x = a

^{7}, then by definition of the order of an element, ord(x) = 3 since x

^{3}= a

^{21}= u. Correct?

By the same reasoning, ord(y) = 7 since y

^{7}= a

^{21}= u.

I would then think that ord(xy) = 1. Since xy = a

^{7}*a

^{3}= a

^{21}= 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 a

^{n}= u, then a

^{pq}= u.

Now if x is an element where x = a

^{p}, then by definition of the order of an element, ord(x) = q since x

^{q}= (a

^{p})

^{q}= a

^{pq}= u.

Similarly, ord(y) = p since y

^{p}= (a

^{q})

^{p}= a

^{qp}= u.

Again, ord(xy) = 1 since xy = a

^{p}*a

^{q}= a

^{pq}= a

^{n}= 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 x

^{x}has infinite order. (Hint: This is a simple proof by contradiction, What happens if (x

^{i})

^{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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