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Show that if $H$ is a subgroup of $(\mathbb Z,+)$ then $\exists n\in\mathbb N$ such that $H=n\mathbb Z$.

And here's what I did:

Suppose that H is a subgroup of Z and $\forall n\in\mathbb N$, $H \neq n\mathbb Z$.

then $\forall n\in\mathbb N$ ($\exists x \in H$ and x $\notin n \mathbb Z $) or ($\exists x \in n\mathbb Z$ and x $\notin H $)

suppose $\forall n\in\mathbb N$ ($\exists x \in H$ and x $\notin n \mathbb Z $)

for n=|x|we have x $\in H$ and x $\in |x|\mathbb Z$ which is absurd.

suppose $\forall n\in\mathbb N$ ($\exists x \in n\mathbb Z$ and x $\notin H $)

since x $\in \cap n \mathbb Z$ then x=0 which is $\in$H and this is absurd.

When I showed this proof to my recitation teacher, he said that the proof is wrong, but he couldn't find why.

user26857
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Guest11235
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  • I found an n such that x is in H and in nZ, since x in H, then x is in Z, it can be either positive or negative, if it is positive then it is an element of xZ. if it is negative then it is an element of -xZ. – Guest11235 Nov 16 '16 at 20:11
  • @DietrichBurde He is arguing by contradiction. He assumes that $x\in H$ and $x\not\in n\mathbb{Z}.$ Then he shows that $x\in |x|\mathbb{Z},$ which gives a contradiction. – mfl Nov 16 '16 at 20:11

3 Answers3

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You wrote "$\forall n\in \Bbb N \, \exists x\in H$" so this $x$ depends on $n$ and it isn't universal which means that for $n,p\in \Bbb N$ there are $x_n, m_p\in H$ where $x_n$ and $x_p$ may be different.

user296113
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The first mistake in your reasoning has been reveiled in @user296113 's answer.

A hint:

If $H\ne\{0\}$ there is a smallest positive element in $H$, call it $n$.

Christian Blatter
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The main problem with your proof is the following. When you say that if $H\neq n\mathbb{Z}$ then $\exists x\in n\mathbb{Z}:x_n\not\in H$ you can't get that conclussion. What you can say is that $\exists x_n\in n\mathbb{Z}:x_n\not\in H.$ And now you can't use that $\cap n\mathbb{Z}=\emptyset$ because it is possible that $x_n\ne x_m.$

mfl
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