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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.

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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.

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

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  • $\begingroup$ oh, I talked this question with my classmate, I think I was midleaded. $\endgroup$ – algebra.And.analysis Sep 26 at 14:20
  • $\begingroup$ A silly question. I don't think it should appear on MSE. I want to delete it... $\endgroup$ – algebra.And.analysis Sep 26 at 14:31
  • $\begingroup$ 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...... $\endgroup$ – Lee Mosher Sep 26 at 15:57

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