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.
Statement
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
.
Proof
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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