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Can anyone please help me prove this 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?

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2 Answers 2

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 $g$.

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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?

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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

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