Question: The exercise is to find the accumulation points of the set $S=\{(\frac {1} {n}, \frac {1} {m}) \space m, n \in \mathbb N\}$

I'm trying to prove that if $A$={accumulation points of the set $S$}, then $A=\{(0,0), (\frac {1} {n},0),(0,\frac {1} {m}) \space n,m \in \mathbb N\}$. I could prove that the set of the right side of the equality is included in $A$. I don't know how to prove the other inclusion, which means that if $x$ is an accumulation point of $S$, then $x$ has to be $(0,0)$, or of the form $(\frac {1} {n},0)$ or $(0,\frac {1} {m})$.

Answer: 
I think you might mean $N\cap Z(G) \neq \{e\}$. In which case,

1. Use the Class equation
2. If $x\in N$, then the conjugacy class $C(x) \subset N$
If $H$,$K$ are any subgroups of a group $G$, with identity $e$, then $H \setminus K \neq \langle e \rangle $, since $e \in H,K$ and hence $e \not\in H \setminus K $. Your question is a special case of that.