Prove that every subgroup of Z is either the trivial group, {0}, or nZ = {nx | x E Z} for some n E N.

Answered: Prove that every subgroup of Z is…

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter4: More On Groups

Section4.5: Normal Subgroups

Problem 27E: 27. Suppose is a normal subgroup of order of a group . Prove that is contained in , the center of...

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Prove that every subgroup of Z is either the trivial group, {0}, or nZ = {nx | x E Z} for some n E N.

Expert Solution

Step 1

To prove:

That every subgroup of $ℤ$ is either the trivial group{0} or $n ℤ = \{n x x \in ℤ\} f o r s o m e n \in ℕ$ .

Proof:

Since,{0} is a trivial group of $ℤ$ ,then there is nothing to prove.

Therefore,It is a subgroup of $ℤ$ .

Now, we need to prove that $n ℤ = \{n x x \in ℤ\} f o r s o m e n \in ℕ$ is the subgroup of $ℤ$ .

a) First,It is closed under addition on $n ℤ$ .

If $a, b \in n ℤ$ , then there exist $k_{1}, k_{2} \in ℤ$ ,such that $a = k_{1} n$ and $b = k_{2} n$ .

Then,

$a + b = k_{1} n + k_{2} n = (k_{1} + k_{2}) n \in n ℤ$

b) The identity of $ℤ$ , 0, is an element of $n ℤ$ , since $0 = n \times 0$ ,so $0 \in n ℤ$ .

(c) Finally, let’s show that any element of $n ℤ$ has an inverse. Indeed if $a \in n ℤ$ , then $a = k_{1} n$ for some integer $k_{1}$ .

Then,

$- a = - k_{1} n = (- k_{1}) n \in n ℤ$ .

Therefore ,the inverse of a is also an element of $n ℤ$ .

Therefore,

from(a),(b) and (c)

$n ℤ = \{n x x \in ℤ\} f o r s o m e n \in ℕ$ is a subgroup of $ℤ$ .

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Prove that every subgroup of Z is either the trivial group, {0}, or nZ = {nx | x E Z} for some n E N. %3D