I don't have any idea where to start.
Please give me hint(s) to prove the following:
Let be a group with be of infinite order.
Let for every positive integer .
If there exists a normal subgroup of finite index in such that for some positive integer , show that .
I'm very sure that it's .
Well, I'm not really understand where to use the 2nd line.
But I did try. Can you please check for me?
Suppose .
Then for some positive integer and .
Since , we let .
Now we have .
This implies .
So, .
Thus contradiction occurs.