Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone please help me prove this question?

Question: So first we can assume that $G$ is not cyclic. Then we need to show that $G$ is a union of proper subgroups. How can I do this?

share|cite|improve this question

HINT: If $G$ is not cyclic, $\langle g\rangle\ne G$ for any $g\in G$. If $G$ is cyclic, and $g$ is a generator of $G$, then $G$ is the only subgroup of $G$ containing $g$.

share|cite|improve this answer

Take any $\,g_1\in G\,$ . Since $\,G\neq\langle g_1\rangle\,$ ,we have that there exists $\,g_2\in G-\langle g\rangle\,$ . So now look at $\,\langle g_1\rangle\cup\langle g_2\rangle\,$ , If this is $\,G\,$ we're done, otherwise there exists $\,g_3\in G-\left(\langle g_1\rangle\cup\langle g_2\rangle\right)\,$ , so look at

$$\bigcup_{i=1}^3\langle g_i\rangle$$

and etc.

Added on request: Suppose

$$G=\bigcup_{H\lneq G} H$$

If $\,G\,$ is cyclic then $\,G=\langle x\rangle\,$ , for some $\,x\in G\,$ , but then

$$\color{red}{\forall\;H\lneq G\;,\;\;x\notin H}\;\Longrightarrow \bigcup_{H\lneq G}H\neq G\,\,\,\,\text{...contradiction!}$$

The gist of the above slick proof is the red part: can you see why it is true?

share|cite|improve this answer
BTW, you can see the construction above has a "little" problem, right? What if the order of $\,G\,$ is way too big ,say uncountable? The construction above still can work but you'll have to go into pretty large ordinals (see e.g. the book in group theory by Kurosh for some very nice examples of this) – DonAntonio Feb 11 '13 at 3:07
And the converse?? – Barbara Osofsky Feb 11 '13 at 6:02
@DonAntonio: Can you please elaborate a bit more on proving that if G is a union of proper subgroups then G is NOT cyclic? – Faye Feb 11 '13 at 7:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.