Prove that a finite group is the union of proper subgroups IFF the group is not cyclic.

Let G be a finite group.

Suppose G is the union of proper subgroups $b_{i}$.

This means that there is an element in G that is not in $b_{1}$. Iterating this reasoning, we see that there is at least one element in G that is not in the union of the proper subgroups $b_{i}$. Then what? This feels like those days where no questions can be solved.

Any hint is appreciated. Thanks in advance.

Edit: If G were cyclic, then, there exists an element, say $a$, that generates G. But $G=\cup _{i=1}^{n}b_{i}$ implies that $a$ generates $\cup _{i=1}^{n}b_{i}$ too. Hence, $\cup _{i=1}^{n}b_{i}$ is not a proper union of subgroup since every element in G is in $\cup _{i=1}^{n}b_{i}$.


2 Answers 2


Assume $G=\displaystyle\bigcup_{i\in I} H_i$, and let any $x\in G$. Then $x\in H_i$ for some $I$ and so $\langle x\rangle \subset H_i$, and so since $H_i$ is proper, $\langle x\rangle $ is as well: $x$ doesn't generate $G$. This is for any $x\in G$, so that $G$ is not cyclic.

Conversely, assume $G$ is not cyclic. Then $G=\displaystyle\bigcup_{x\in G}\langle x\rangle$ is a union of proper subgroups.

  • $\begingroup$ Is the element x necessarily restricted to being in one proper subgroup or can it be contained in the intersection of at least 2 proper subgroup? $\endgroup$ Commented Apr 15, 2017 at 10:03
  • $\begingroup$ $x$ can be in many subgroups, that doesn't change anything $\endgroup$ Commented Apr 15, 2017 at 10:55

Let $G$ be a group, finite or infinite. Observe that the following statements are equivalent.

  1. $G$ is the union of some proper subgroups.
  2. $G$ is the union of all of its proper subgroups.
  3. Each element of $G$ belongs to some proper subgroup of $G.$
  4. For each element $g\in G,$ $\langle g\rangle$ is a proper subgroup of $G.$
  5. There is no element $g\in G$ such that $\langle g\rangle=G.$
  6. $G$ is not cyclic.

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