Proving a subgroup of the symmetric group $(S_z, ∘)$

Question

$S_z = \{F: Z → Z : F $ is a bijection onto$\ Z\}$. $(S_z, ∘)$ is a group.

Question

Let $H = \{F ∈ S_z : $ for each n$\ ∈ Z_+, F(n) ∈ Z_+\}$. Is $H$ a subgroup of $(S_z, ∘)$?

My Proof:

$H ≠ ∅$ as the identity function $i ∈ S_z ∈ H$, since $i(n) = n$ for all $n ∈ F(n)$.

Let $f, g ∈ H$, then $f ∘ g(n) = f(g(n)) = f(n) = n$.

If $f ∈ S_z$, then $f^{-1}(n) = f^{-1} (f(n)) = i(n) = n$, since $f(n) = n$. Therefore, $f^{-1} ∈ H$.

Since we have proven the identity function exists, closure, and the inverse function exists, H is a subgroup.

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I feel like I have a huge misunderstanding on what the set H is. In my mind it's $\{1, 2, ..., n\} → \{x_1, x_2, ..., x_n\} $ where$\ n ∈ Z_+$.

(I'm not very good at solving these problems)

Answer 1

If it's a subgroup, every element should have an inverse. The problem here is that there are bijections from Z to Z which send Z+ into Z+ but that don't have similar inverses. For example, send every number x to x+1. This is a bijection and all the positives go to positives, but you can't send 1 back to 0 and stay in the positives. So no... it's not a subgroup.

Answer 2

Let $f:\mathbb{Z}\to\mathbb{Z}$ be defined by $f(n)=n+1$, for all $n\in \mathbb{Z}$.

Then $f\in H$, but $f^{-1}\notin H$, hence $H$ is not a subgroup of $S_{\mathbb{Z}}$.