# Thread: proof for multiplicative subgroups

1. ## proof for multiplicative subgroups

I don't have any idea where to start.
Please give me hint(s) to prove the following:

Let $\displaystyle A$ be a group with $\displaystyle h \in A$ be of infinite order.
Let $\displaystyle x \in A \backslash \langle h^{s} \rangle$ for every positive integer $\displaystyle s$.
If there exists a normal subgroup of finite index $\displaystyle N$ in $\displaystyle A$ such that $\displaystyle N \cap \langle h \rangle=\langle h^{rs} \rangle$ for some positive integer $\displaystyle r$, show that $\displaystyle x \notin \langle h^{s} \rangle N$.

2. Originally Posted by deniselim17
I don't have any idea where to start.
Please give me hint(s) to prove the following:

Let $\displaystyle A$ be a group with $\displaystyle h \in A$ be of infinite order.
Let $\displaystyle x \in A \backslash \langle h^{s} \rangle$ for every positive integer $\displaystyle s$.
If there exists a normal subgroup of finite index $\displaystyle N$ in $\displaystyle A$ such that $\displaystyle N \cap \langle h \rangle=\langle h^{rs} \rangle$ for some positive integer $\displaystyle r$, show that $\displaystyle x \notin \langle h^{s} \rangle N$.

Surely you meant $\displaystyle x\notin h^sN$ in your last line above...

Sime ideas: $\displaystyle x\in h^sN\Longleftrightarrow x=h^sn\,,\,\,n\in N$ . Now, suppose $\displaystyle [G:N]=k\Longrightarrow g^k\in N\,\,\forall g\in G$ , and

thus $\displaystyle x^k\in N\,,\,say\,\,x^k=n_1\,,\,n_1\in N\Longrightarrow x^k\in \left(h^s N\right)^k=h^{ks} N\Longrightarrow n_1=h^{ks}m\,,\,m\in N$ ...

Tonio

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

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