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Let $\Bbb C^*$ be the multiplicative group of complex numbers. Consider the subgroup $$U=\{z \in \Bbb C~:~z^n=1,~n=1,2,3,...\}.$$ Interestingly, the group $U$ is an infinite group with each element has a finite order. However, $U$ has an infinite proper subgroup $$G=\{z \in \Bbb C~:~z^{2^n}=1,~n=0,1,2,3,...\}.$$

Let us denote the cyclic group $\{z \in \Bbb C~:~z^n=1\}$ by $U_n$ to say $$U=\bigcup_{n=1,2,..}U_n, ~\text{and}~G=\bigcup_{n=0,1,2,..}U_{2^n}.$$

I am sure that, ''any proper subgroup of $G$ is finite and cyclic unlike that of $U$'', But I want to justify it.

Trying for $H=\{z \in \Bbb C~:~z^{4^n}=1,~n=0,1,2,3,...\}$ was a feeble attempt as I realised that $H=G$.

I hope the fact ''$G$ is a big union of nested cyclic groups'' is useful to prove the claim. But how to?

thanks in advance.

Messi Lio
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1 Answers1

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Note that all elements of $G$ have a finite order which is a power of $2$.

Let $g$ and $h$ be elements of $G$ with $ord(g)=2^a$ and $ord(h)=2^b$. If $a\leq b$, let’s show that $g$ is a power of $h$.

By the Fundamental Theorem of Algebra, we have that $x^{2^b}=1$ has exactly $2^b$ solutions and $h,h^2,h^3,...,h^{2^b}$ are $2^b$ distinct values which satisfy the equation, so they are all the solutions to the equation. Since $g^{2^b}=(g^{2^a})^{2^{b-a}}=1$, $g$ must be in that set and thus a power of $h$ as desired.

Note that in particular, this implies that every two elements of $G$ have one being a multiple of the other (or both).

Let H be a subgroup of G.

Let $S=sup_{g\in H} ord(g)$.

If $S$ is finite, then every element of $h\in H$ has order at most $S$. Since all the orders are integers, the supremum is obtained by some element $g\in G$ with order $S$. Then, having the highest order, every other element of $H$ is a power of $g$, so $H$ is cyclic of order $S$.

Otherwise, $S$ is infinite. Let $g\in G$ have some order $2^a$. Then, there’s some element $h$ in $H$ with greater order $2^b \geq 2^a$. By the above, $g$ must be a power of $h$ and so in $H$. Thus, every element of $G$ is in $H$ which is a subset of $G$, so $G$ itself is $H$.

Thus, any subgroup of $G$ either is cyclic of finite size or is $G$ itself.

Eric
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  • How to say ''There are only $S(S+1)/2$ such values'' satisfying at least one of $x^1=1,x^2=1,...,x^S=1$? – Messi Lio May 15 '23 at 06:18
  • OR in particular, how can we say ''there are only 3 values (beyond $1$ and $-1$) satisfying at least one of $x=1,x^2=1$''? – Messi Lio May 15 '23 at 06:22
  • I added more detail including showing it’s cyclic. The key bit is that $x^a=1$ has exactly $a$ solutions. – Eric May 15 '23 at 07:04