Show ZXZ/<1,1> is an infinite cyclic group.

In summary, ZXZ/<1,1> is an infinite cyclic group generated by <1,0> or (1,0) and is isomorphic to Z+0Z, which is also infinite and cyclic. The factor group ZXZ/<1,1> is represented by Z+<1,1> and all elements can be generated by <1,0> or (1,0). The identity element is (0,0) and (1,0) is the generator. Therefore, ZXZ/<1,1> is a cyclic infinite group.
  • #1
Daveyboy
58
0

Homework Statement



Show ZXZ/<1,1> is an infinite cyclic group.

Homework Equations





The Attempt at a Solution


<1,1> = {...(-1,-1), (0,0), (1,1),...}

implies ZXZ/<1,1> = {<1,0>+<1,1>, <0,1>+<1,1>} which is isomorphic to ZXZ.

But ZXZ is not cyclic, is my description of the factor group wrong , is it not isomorphic to ZXZ, or am I missing other points?
 
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  • #2
what characterize an infinite cyclic group?

1. generated by a single element
2. infinite order

can you put an element canonically into a certain form?
so say you have (m, n), what can I write it as?
 
  • #3
if (m,n) is an element of one of the left cosets of the factor group then,
if m=n (m,n) is in <1,1>. if m/=n then (m,n) = (m,n) + <1,1>. (because (0,0) is
generated by <1,1>). I'm not sure what this implies though.
 
  • #4
so actually <1,0> + <1,1> generated all elements of ZXZ. Is it acceptable to say ZXZ/<1,1> is isomorphic to Z+0Z which is isomorphic to Z? (Which is generated by <1> or <-1>)
 
  • #5
<1,1> is just the identity. So, yeah.
 
  • #6
... this is the additive group, (0,0) is the identity, and I'm using the notation <1,1> to imply that this is the cyclic generator. I hope that clears up any confusion.

I'm not sure if the conclusion I've come to is correct.

ZXZ/<1,1> = {Z+<1,1>}
Clearly Z is infinite and cyclic generated by <1> or <-1>.
So, ZXZ/<1,1> is a cyclic infinite group.
 
  • #7
well, (0,0)=(1,1) since you are modding out the diagonal set.
(1,0) is the generator (or using coset notations (1,0)+<(1,1)>). Your conclusion is definitely correct (as in post 4)
 
  • #8
awesome thanks.
 

Related to Show ZXZ/<1,1> is an infinite cyclic group.

1. What is an infinite cyclic group?

An infinite cyclic group is a mathematical structure that consists of a set of elements and a binary operation (usually denoted as *) that satisfies the following properties:

  • The set is infinite, meaning it has an infinite number of elements.
  • The binary operation is associative, meaning (a*b)*c = a*(b*c) for all elements a, b, and c in the group.
  • There exists an identity element e in the group, such that a*e = e*a = a for all elements a in the group.
  • For each element a in the group, there exists an inverse element a^-1, such that a*a^-1 = a^-1*a = e.
  • The group is generated by a single element, meaning all other elements can be obtained by repeatedly applying the binary operation to this element.

2. How is "Show ZXZ/<1,1> is an infinite cyclic group" related to infinite cyclic groups?

The statement "Show ZXZ/<1,1> is an infinite cyclic group" means to prove that the quotient group ZXZ/<1,1> is an infinite cyclic group. In this context, ZXZ refers to the set of all integer matrices of the form [a b; c d], where a, b, c, and d are integers. The binary operation in this group is matrix multiplication. The subgroup <1,1> refers to the set of all matrices of the form [1 0; k 1], where k is an integer. The quotient group ZXZ/<1,1> is the set of all cosets (equivalence classes) of ZXZ under the subgroup <1,1>. Therefore, proving that ZXZ/<1,1> is an infinite cyclic group means showing that all elements in this quotient group can be generated by a single element, making it an infinite cyclic group.

3. What is the significance of proving that ZXZ/<1,1> is an infinite cyclic group?

Proving that ZXZ/<1,1> is an infinite cyclic group is significant because it helps us understand the structure and properties of this group. As an infinite cyclic group, ZXZ/<1,1> has many interesting mathematical properties, such as being isomorphic to the group of integers under addition. This knowledge can be applied in various areas of mathematics, including number theory, algebra, and geometry.

4. How can one prove that ZXZ/<1,1> is an infinite cyclic group?

There are several ways to prove that ZXZ/<1,1> is an infinite cyclic group. One approach is to show that every element in this quotient group can be generated by a single element, such as [1 0; 1 1]. This can be done by showing that any matrix in ZXZ/<1,1> can be expressed as a power of [1 0; 1 1]. Another approach is to show that ZXZ/<1,1> is isomorphic to another known infinite cyclic group, such as the group of integers under addition. This can be done by constructing a bijective homomorphism between the two groups.

5. Can ZXZ/<1,1> be generated by any other element besides [1 0; 1 1]?

Yes, ZXZ/<1,1> can be generated by other elements besides [1 0; 1 1]. For example, the element [2 0; 2 2] can also generate all other elements in this quotient group. However, [1 0; 1 1] is a convenient choice for generation as it has a simple form and can be easily expressed as a power of itself. It is also the smallest positive integer matrix that can generate all other elements in ZXZ/<1,1>.

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