Find an abelian infinite group such that every proper subgroup is finite
I've seen the answer here but I'm really struggling to understand and prove it.
What I understand is that if I suppose that $H$ is infinite, then for all $n$, We can find $z$ such that $z$ generates $\Bbb{U_{2^p}}$ where $p > n$ (why?) and $\Bbb{U_{2^n}}$ is in $\Bbb{U_{2^p}}$
Can someone explain it a bit more?
Think of a pizza. Cut it in half. Those endpoints of the diameter are the square roots of unity. Cut the halves in half. Those endpoints are the additional (primitive) $4$th roots of unity, although the previous points are also $4$th roots of unity. Cut those $4$ pieces in half to produce $4$ additional (again, primitive) $8$th roots of unity, but all the previous $4$ were also $8$th roots of unity. Et cetera.
In summary, all square roots of unity are $4$th roots of unity, which are themselves $8$th roots of unity, etc.
Why? If $z^2 = 1$, then $z^4 = (z^2)^2 = 1^2 = 1$, and $z^8 = (z^4)^2 = 1^2 = 1$, etc.
In general, if $p>n$, then a root of unity $z$ of order $2^n$ is automatically a root of unity of order $2^p$ by the same calculation: $$ z^{2^p} = \bigl( z^{2^n} \bigr)^{2^{p-n}} = 1^{2^{p-n}} = 1, $$ which is meaningful because $p-n>0$.
Aside. This fact is even more general (but not useful in this particular exercise): If $m$ and $n$ are naturals and $m \mid n$, then an $m$th root of unity is automatically an $n$th root of unity. Why? Write $n = mk$ for some $k \in \mathbb{N}$. Then, $$ z^n = z^{mk} = \bigl( z^m \bigr)^k = 1^k = 1. $$
