I'm studying this weird course called "Analytic Geometry", but in reality it seems like a mash of modern or abstract Algebra (...I'm not so sure...), and includes stuff like Affine transformations, isomorphism, quadrics etc'. It's a really weird course and I can't find any legit source of knowledge to make an order in all the mass.
Anyway, we were introduced (kinda..) to the concept of groups. Defining it: Set $G$ of functions $f:x \rightarrow x$ is called a group if:
a) Identity function $Id_x:x\rightarrow x \in G$
b) If $f,g \in G$ then also $f \circ g \in G$ and $g \circ f \in G$.
c) Every $f \in G$ has an inverse.
Anyway, I was given the following function group $N_a: \mathbb Z\rightarrow\mathbb Z$ defined $N_a(x)=2a-x$. And so I was asked whether the group $\{N_a|a\in \mathbb Z \}$ is a group. And I got quite confused trying to solve it. Thanks in advance for any assistance!
Q:Is the set $G=\{N_a\mid a\in\mathbb Z\}\neq\emptyset$?A:Yes. For example $N_1\in G$ when $a=1$.Q:Does $N_a,N_b\in G$ when $a,b\in\mathbb Z$ lead us to have $N_a\circ N_b\in G$?A:Let's find that composition for arbitrary $z\in\mathbb Z$. $$N_a\circ N_b(z)=N_a(2b-z)=2a-(2b-z)=2(a-b)+z$$ We see that the result is not as we have expected, because it should have been $2k-z$ for some integer $k$. So we stop because we see that we need additional restrictions in the body which are not given. Therefore checking other points of definition of a group is not necessary. However you may see that $G$ has no element such that $N_a(x)=x$.