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More clearly, my question is:

Let $G$ be a finite group, there are some cyclic subgroups $H_i$ of $G$,

$\text{s.t. } G=\bigcup H_i,H_i \cap H_j = \{ e \}(i \ne j)$.

I know if $G$ is infinite, it is wrong. (I consider the example $(\mathbb Z,+)$)

I believe when $G$ is finite, it is wrong, too. But I find it difficult for me to construct counter examples.

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  • $\begingroup$ You've got to be careful about a union of subgroups: the union won't necessarily be a subgroup itself. For your counterexample, the Klien-4 group has two cyclic copies of $\mathbf{Z}_2$, and is isomorphic to $\mathbf{Z}_2 \times \mathbf{Z}_2$, but it's certainly not equal to the union of those two copies of $\mathbf{Z}_2$. That union would only have three elements, and it isn't a subgroup. $\endgroup$ – Mike Pierce May 30 '18 at 15:24
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    $\begingroup$ Quaternion group? $\endgroup$ – Lord Shark the Unknown May 30 '18 at 15:24
  • $\begingroup$ @Mike Pierce Thanks but I don't understand you. I think Klien-4 group has three cyclic copies of Z2, because the square of every element of the Klien-4 group is e. $\endgroup$ – namasikanam May 30 '18 at 15:48
  • $\begingroup$ Yeah, you're right. I misunderstood your question. $\endgroup$ – Mike Pierce May 30 '18 at 15:57
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Thanks to @Lord Shark the Unknown.

A quaternion group is exactly a counter example.

A quaternion group has only 5 cyclic subgroup:

$H_1=\{ 1 \}\\H_2=\{ 1, -1 \}\\H_3=\{1, i, -1, -i \}\\H_4=\{1, j, -1, -j\}\\H_5=\{1, k,-1 ,-k\}$

So if let the union is $G$, then $i,j \in G$. We must choose $H_3$ and $H_4$, but $H_3 \cap H_4 \not= \{ 1 \}$.

I do not really understand what a quaternion group really is, but I believe it very beautiful.

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    $\begingroup$ Quaternion group of order 8, $Q_8$ is indeed beautiful. It also serves as a counter-example for the converse of the fact that "every subgroup of an abelian group is normal". If you observe subgroups of $Q_8$, you will notice that all of them are normal, but $Q_8$ is not abelian. $\endgroup$ – feynhat May 30 '18 at 17:32

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