Tour:Every nontrivial subgroup of the group of integers is cyclic on its smallest element
This article adapts material from the main article: every nontrivial subgroup of the group of integers is cyclic on its smallest element
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WHAT YOU NEED TO DO:
- Read the statement and understand it thoroughly.
- Go through the proof very carefully. Make sure you understand all the steps of the proof, particularly the application of Euclidean division.
Let be a subgroup of , the group of integers under addition. Then, there are two possibilities:
- is the trivial subgroup, i.e.
- contains a smallest positive element, say , and is the set of multiples of . Thus, is an infinite cyclic group generated by , and is isomorphic to . We typically write .
Given: A nontrivial subgroup of , the group of integers under addition
To prove: There exists a smallest positive element in , and , so is isomorphic to
Proof: First, observe that if is nontrivial, then there exists a nonzero element in . This element may be either positive or negative. However, since is a subgroup, it is closed under taking additive inverses, so even if the element originally picked was negative, we have found a positive number in .
Thus, the set of positive numbers in is nonempty. Hence, there exists a smallest positive number in . Call it .
Clearly, all the integer multiples of are in . We need to prove that every element in is a multiple of .
By the Euclidean division algorithm, we can write:
where are integers and . Since , . Thus, is a nonnegative integer less than such that . By the minimality of , we have , so , as desired.
Thus, , or is the set of multiples of .
An explicit isomorphism from to is given by the map sending an integer to the integer .
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