Originally Posted by

**deniselim17** I'm very sure that it's $\displaystyle x \notin \langle h^{s} \rangle N$.

Well, I'm not really understand where to use the 2nd line.

But I did try. Can you please check for me?

Suppose $\displaystyle x \in \langle h^{s} \rangle N$.

Then $\displaystyle x=h^{st}n$ for some positive integer $\displaystyle t$ and $\displaystyle n \in N$.

Since $\displaystyle x \in \langle h \rangle$

Why? From where did you conclude this?

Tonio

, we let $\displaystyle x=h^{p}, p\in \mathbb{Z}$.

Now we have $\displaystyle x=h^{p}=h^{st}n$.

This implies $\displaystyle n \in N \cap \langle h \rangle = \langle h^{rs} \rangle$.

So, $\displaystyle x=h^{p}=h^{st}h^{\alpha rs}=h^{s(t+\alpha r)} \in \langle h^{s} \rangle$.

Thus contradiction occurs.