# On the group of all complex roots of unity whose orders are powers of $p$, prime number

$A=\{z\in\mathbb{C}: z^{p^n}=1\;\mathrm{for\;some\;integer}\;n\geq1\}$. I have to prove:

1) Every proper subgroup of $A$ is cyclic,

2) if $B,C$ are subgroups of $A$ then $B\subseteq C$ or $C\subseteq B$,

3) for every $n\geq0$ there exists an unique subgroup of $A$ with $p^n$ elements.

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The group is also called the Prufer $p$-group. There have been lots of questions on this group on the website before. See here, and here, for example. – Arturo Magidin Jan 2 '12 at 7:05

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.
2. 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 $B\subseteq C$.
3. This is simply the group of $p^n$-th roots of unity. Again, the same argument shows that this is unique.
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.