# Question about a group which can be decomposed as the union of all its proper subgroups.

This question is from prufer group.

Suppose $$G$$ is a group. I am trying to find an example about $$G = \bigcup\limits_{α \in I}A_α$$, where $$\{A_α \mid α \in I\}$$ is the set of all proper subgroups of $$G$$ and $$A_α\cap A_β = \{1\}$$ whenever $$α\neq β$$.

According to sylow theorem, it is easy to know that $$G$$ can not be finite group.

If there is an element $$a$$ of infinite order, then $$(a)$$ is an infinite cyclic subgroup of $$G$$. Thus $$(a^2) \cap (a) \neq \{e\}$$.

Consequently, $$G$$ must be an infinite group and the order of every element in $$G$$ is prime.

You can classify the finite groups satisfying this: using Sylow's theorems, you can winnow the possibilities for the order of the group; for instance, if the order has a factor of the form $$qp^2$$ for not-necessarily distinct primes $$q$$ and $$p$$, then it has a subgroup of order $$p^2$$ which has a subgroup of order $$p$$ - which would contradict the condition. Thus, the order must either be the square of a prime or square-free.

For the first case, the only groups of order $$p^2$$ are $$C_{p^2}$$ and $$C_p\times C_p$$ where $$C_n$$ is a cyclic group of order $$n$$. The group $$C_p\times C_p$$ satisfies the property and $$C_{p^2}$$ does not.

The square-free case leads to more examples. If you have just two factors, you find that every non-abelian group of order $$pq$$ has the desired property (and that these groups exist exactly when $$q|p-1$$ and are given by semidirect products) - so, for instance, the dihedral groups of order $$2p$$ are examples.

There are no other examples of finite groups with this property. Let $$G$$ be any group with square-free order of order $$n=p_1p_2\ldots p_k$$ for $$k$$ distinct primes $$p_1,\ldots,p_k$$. You could find subgroups $$P_1,P_2,P_3,\ldots$$ of those given orders. Note that the normalizer of any non-trivial subgroup $$S$$ has to either be that subgroup itself or the entire group, since otherwise one would have $$0 < S < N_G(S) < G$$ which would contradict the condition. We must not have that $$N_G(P_i)=G$$, since then $$P_i$$ would be normal and $$G/P_i$$ would be a group of non-prime order, hence would have a proper non-trivial subgroup, the preimage of which under the quotient map would be a proper subgroup of $$G$$ strictly containing $$P_i$$ - which again, would fail the condition. At this point, we run into a simple counting problem: each group $$P_i$$ must have $$\frac{n}{p_i}$$ conjugates which intersect trivially pairwise - leading to $$n\left(1-\frac{1}{p_i}\right)$$ elements of order $$p_i$$ in $$G$$ for each index $$i$$. There are not enough elements in $$G$$ to satisfy this for each prime - directly, we have claimed there to be $$n\left(2-\frac{1}{p_1}-\frac{1}{p_2}\right)$$ of order either $$p_1$$ or $$p_2$$, but this quantity exceeds $$n$$, so this is a contradiction.

There are also infinite groups with this property, for instance Tarski monster groups are all examples of the property you ask for.

You say that $$G$$ cannot be finite. But what about $$G = \mathbb Z / 2 \mathbb Z \oplus \mathbb Z / 2 \mathbb Z$$?

• oh, I talked this question with my classmate, I think I was midleaded. – algebra.And.analysis Sep 26 at 14:20
• A silly question. I don't think it should appear on MSE. I want to delete it... – algebra.And.analysis Sep 26 at 14:31
• I would not say it is silly. I actually had to think about this for a little while! And it made me think about followup questions too...... – Lee Mosher Sep 26 at 15:57