If $C$ is a cyclic subgroup of a group $G$, let $gen(C)$ denote the set of all its generators. Then $G$ is the disjoint union $$G = \cup gen(C)$$ where $C$ ranges over all the cyclic subgroups of $G$.

Rough proof:
Suppose $G=1$. Then we are done.
Otherwise, $\exists b\in G\backslash 1$.
Since $\langle b \rangle \neq 1$ and $\langle b \rangle \leq G$.
If $G=1 \sqcup gen(b)$, then we are done.
Otherwise,$\exists c\in G\backslash(1 \sqcup gen(b))$
Then $\langle c \rangle \neq \langle b \rangle,\langle c \rangle\neq 1$ and $\langle c \rangle \leq G$.
If $G=1 \sqcup gen(b) \sqcup gen(b)$, then we are done. Otherwise, we repeat the same manner.

I am not sure that my proof is correct or not.


I would say that the main idea of the proof is correct, and it can be made rigorous (using induction argument) if the group $G$ is finite. But what if $G$ is infinite? I am guessing that you can still use something like Zorn's lemma to make it rigorous.

I would suggest the following simpler argument that works for all groups $G$ (finite or infinite). Let $a\in G$. Then $C=\langle a\rangle$ is a cyclic subgroup of $G$, and by definition $a$ generates it, so $a\in \operatorname{gen}(C )$. Since $a\in G$ was an arbitrary element, this shows that $$ G\subseteq \bigcup_{C\text{ cyclic}} \operatorname{gen}(C) $$ But the reverse inclusion $$ G\supseteq \bigcup_{C\text{ cyclic}} \operatorname{gen}(C) $$ is clear, since everything lives inside $G$. Putting these inclusions together, you get the desired equality.

Perhaps this is clear to you, but it is worth thinking why the union $\bigcup_{C\text{ cyclic}} \operatorname{gen}(C )$ is a disjoint union. Namely, if $C_1$ and $C_2$ are two different cyclic subgroups, then they cannot have common generators.

  • $\begingroup$ Can i say suppose $\langle c_1 \rangle = C_1\neq C_2 = \langle c_2 \rangle$. Then $\exists x\in C_1 \backslash C_2$. Suppose $c_1=c_2$. Then $x=c_1^j=c_2^j \in C_2$, which is a contradiction? $\endgroup$ – Alan Wang Jan 25 '16 at 6:59
  • $\begingroup$ @Alan Wang: Yes, exactly! $\endgroup$ – Prism Jan 25 '16 at 17:44

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