The multiplicative group of all complex $2^n$-th roots of unity, where $n = 0 , 1, 2 , \ldots$ [duplicate]
Solution 1:
The key here is to observe that a proper subgroup $B\subset A$ must be missing some primitive $p^m$-th root of unity, hence must miss all $p^k$-th roots of unity for $k \geq m$ (as otherwise we could simply take products of this root to get the primitive $p^m$-th root of unity). So there is some $n$ such that the $p^n$-th roots of unity are in $B$ but this is not true for any larger value of $n$. Since a $p^{n-i}$-th root of unity is also a $p^n$-th root of unity, this makes $B$ the group of $p^n$-th roots of unity. It's not hard to show that $B$ must then be cyclic.
A similar argument as for 1 suffices. $B$ must consist of $p^n$-th roots of unity, while $C$ must consist of $p^m$-th roots of unity, and depending on whether $n\geq m$ or $m\geq n$ this means $C\subseteq B$ or $B\subseteq C$.
This is simply the group of $p^n$-th roots of unity. Again, the same argument shows that this is unique.
Solution 2:
Observe that if $B$ is a subgroup of $A$ containing an element a primitive $p^{k}$-th root of unity, then every $p^{k}$-th root of unity is in $B$. So if $B$ contains a primitive $p^{k}$-th root of unity for arbitrarily large $k$, then $B=A$; otherwise, take $k$ to be maximal and $B$ is simply the multiplicative group of $p^{k}$-th roots of unity.