proof for multiplicative subgroups

**deniselim17** · November 9th, 2010, 10:25 PM

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

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

**tonio** · November 10th, 2010, 01:13 AM

Originally Posted by deniselim17

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

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

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

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

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

Tonio

**deniselim17** · November 10th, 2010, 04:51 PM

I'm very sure that it's $x \notin \langle h^{s} \rangle N$ .
Well, I'm not really understand where to use the 2nd line.
But I did try. Can you please check for me?

Suppose $x \in \langle h^{s} \rangle N$ .
Then $x=h^{st}n$ for some positive integer $t$ and $n \in N$ .
Since $x \in \langle h \rangle$ , we let $x=h^{p}, p\in \mathbb{Z}$ .
Now we have $x=h^{p}=h^{st}n$ .
This implies $n \in N \cap \langle h \rangle = \langle h^{rs} \rangle$ .
So, $x=h^{p}=h^{st}h^{\alpha rs}=h^{s(t+\alpha r)} \in \langle h^{s} \rangle$ .
Thus contradiction occurs.

**tonio** · November 10th, 2010, 06:34 PM

Originally Posted by deniselim17

I'm very sure that it's $x \notin \langle h^{s} \rangle N$ .
Well, I'm not really understand where to use the 2nd line.
But I did try. Can you please check for me?

Suppose $x \in \langle h^{s} \rangle N$ .
Then $x=h^{st}n$ for some positive integer $t$ and $n \in N$ .
Since $x \in \langle h \rangle$

Why? From where did you conclude this?

Tonio

, we let $x=h^{p}, p\in \mathbb{Z}$ .
Now we have $x=h^{p}=h^{st}n$ .
This implies $n \in N \cap \langle h \rangle = \langle h^{rs} \rangle$ .
So, $x=h^{p}=h^{st}h^{\alpha rs}=h^{s(t+\alpha r)} \in \langle h^{s} \rangle$ .
Thus contradiction occurs.

.

**deniselim17** · November 10th, 2010, 06:51 PM

Originally Posted by tonio

.

Sorry. There's a typo.
$x \in \langle h \rangle \backslash \langle h^{rs} \rangle$ .

I think my proof works.

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