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Say $F$ is the group of all functions $f: \mathbb Z \rightarrow \mathbb Z$. The group operation is $+$ defined by $(f_1 + f_2)(x) = f_1(x) + f_2(x)$. How would I show that $N = \{ f \in F | f(3) = 0\}$ is a normal subgroup of $F$?

I was thinking of showing that $N$ is the kernel of some homomorphism $\phi: F \rightarrow \mathbb Z$ since the kernel of a homomorphism is always a normal subgroup. However, I could not think of a homomorphism that could do this. Is there a fast way to come up with homomorphisms that have a specific kernel?

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    $\begingroup$ How about evaluation at $3$? That is, $f \in F$ gets sent to $f(3)$. (Incidentally, $F$ is an abelian group, so every subgroup is normal; it therefore suffices to show that $N$ is a subgroup, which isn't hard to show. Nevertheless, these ''evaluation'' homomorphisms are quite common, so it's good to know this sort of example.) $\endgroup$ – Alex Wertheim Nov 21 '16 at 3:19
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    $\begingroup$ $\phi(f) = f(3) $ so that $\phi(f_{1} + f_{2}) = (f_{1}+f_{2})(3)=f_{1}(3) + f_{2}(3) = \phi(f_{1}) + \phi(f_{2}) $ will this not work? $\endgroup$ – emmy Nov 21 '16 at 3:23
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    $\begingroup$ @PiccolMan: happy to help. Yes, that is correct. $\endgroup$ – Alex Wertheim Nov 21 '16 at 3:24
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    $\begingroup$ Yes. It's even simpler - you can consider the constant functions $f_{c} \colon \mathbb{Z} \to \mathbb{Z}$ which send $x$ to $c$ for every $x \in \mathbb{Z}$. $\endgroup$ – Alex Wertheim Nov 21 '16 at 3:51
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    $\begingroup$ $F$ is Abelian. All subgroups are normal. Just show that $N$ is a subgroup (and that's pretty easy!) You don't need to construct a homomorphism to solve this problem. $\endgroup$ – Scott Burns Nov 21 '16 at 4:20
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As it has been pointed out there are several ways. I'll show both.

As it is an abelian group all subgroups are normal. And showing that $N$ is a subgrou is trivial. let $f,g\in N$ then we have $$(f-g)(3)=f(3)-g(3)=0-0$$ So by the subgroup test we have that it is a subgroup.

For a kernel we have for any function group (or ring) the evaluation homomorphism. $\varphi_x(f)=f(x)$ and then we have $\varphi_3:N\to\Bbb Z$ which has the kernel you desire and as such $\ker\varphi_3=N$

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