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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!

mathematics2x2life
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oey
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  • Have you tried to examine the correctness of points above? – Mikasa Dec 22 '13 at 16:07
  • Yeah I'm not sure about the first point.... – oey Dec 22 '13 at 16:10
  • Can you find an $a$ such that $N_a$ equals the identity? If not then you are ready. – drhab Dec 22 '13 at 16:11
  • Can I take $a=x$? Is that allowed..? – oey Dec 22 '13 at 16:20
  • No. $a$ is the constant that determines the function $N_a$. $x$ is the argument of the function. – drhab Dec 22 '13 at 16:31
  • So there is no $a$ so that $N_a$ is the identity. But how can I prove that? – oey Dec 22 '13 at 16:31
  • @oey: If $N_a(x)=x$ then $2a-x=x$ then $a=x$. So the identity element seems to be $N_x(x)$. But by assuming this, we intuitively accept that the identity element is not unique. This is a real problem for the set. – Mikasa Dec 22 '13 at 16:34

2 Answers2

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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 $N_a(x)=x$.

Mikasa
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If $\left\{ N_{a}\mid a\in\mathbb{Z}\right\} $ equipped with composition is indeed a group then $N_{a}=id_{\mathbb{Z}}$ for some $a$. However $N_{a}(x)=2a-x$ so $N_{a}=id_{\mathbb{Z}}$ leads $x=a$ for each $x$ (as a consequence of $2a-x=x$) which is not true. We conclude that this is not a group.

drhab
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