let $G$ be a finite abelian group such that it contains a subgroup $H_0\neq \{e\}$ which is contained in every subgroup $H\neq \{e\}$ of $G$

Prove that $G$ is cyclic.Find $o(G)$

How should I start?

My thoughts: Since $G$ is a finite abelian group ,$G$ can be expressed as a direct product of cyclic groups.Let $G=\mathbb Z_{p_1^{\alpha_1}}\times \mathbb Z_{p_2^{\alpha_2}}\times ...\times\mathbb Z_{p_n^{\alpha_n}}$.

Let $H_0$ be a subgroup of $G$ contained in every other subgroup of $G$ .Now each $\mathbb Z_{p_i^{\alpha_i}}$ can be viewed as a subgroup of $G$ .

So $H_0$ is contained in each $\mathbb Z_{p_i^{\alpha_i}}$ for each $i=1,2,...,n$. So $o(H_0)$ divides $p_i^{\alpha_i}$ for each $i=1,2,...,n$ which is not possible as $p_i$ 's are mutually prime; hence contradiction

So $i=1$ only so $G=\mathbb Z_{p_1^{\alpha_1}}$ which is cyclic.Is this true??Please help.


Suppose $K\le G$ is a non-trivial subgroup of $G$ and that for every non-trivial subgroup $X\le G$, we have $K\le X$. Then $K$ itself has no nontrivial subgroup. Moreover, for any $k\in K\setminus\{1\}$, we have $\langle k\rangle=K$ and thus $K\cong \mathbb Z/p\mathbb Z$ for some prime$p$

Let $g\in G\setminus\{1\}$ be arbitrary. Then $k\in\langle k\rangle = K\le \langle g\rangle$, i.e. $k=g^n$ for some $n$. If we write $n=p^mr$ with $r$ not divisible by $p$, then $h:=g^{p^{m+1}}$ has order dividing $r$ (because $h^r=g^{pn}=k^p=1$) and thus $k\notin \langle h\rangle$, from which we conclude $h=1$, hence $g^{p^{m+1}}=1$ and the order of $g$ is a power of $p$. This shows that $G$ is a $p$-group.Now can you conclude?

  • $\begingroup$ Why dont you say whether mine is correct or not? $\endgroup$ – Learnmore May 17 '15 at 6:49
  • $\begingroup$ @learnmore clearly your proof is not valid as explained by others too. Most of the time I avoid nontrivial theorems and the theorem you used is nontrivial result.(atleast to me). $\endgroup$ – Arpit Kansal May 17 '15 at 16:03

Your proof is not correct.

The main reason it is not correct is that it is not true that the $p_i$ can be taken to be distinct in the decomposition you mentioned. (Moreover it is cleaner to say $G$ is isomorphic to that group, not that it is equal.)

However, you can use the decomposition you recalled and just say that since you have the subgroups you mention and the intersection of two such subgroups is the trivial group there can be only one such group. So, $n=1$ and the group is cyclic. In fact it is even cyclic of prime power order, which is perhaps what is meant by finding the order. (Note that a cyclic group not of prime power order does in fact not have the property you want.)

  • $\begingroup$ The groups will not be the same, not even isomorphic. Note $Z_p \times Z_p$ is not the same as $Z_{p^2}$. (Indeed, when the primes are distinct you could collect then together in one cyclic group $Z_p \times Z_q$ is isomorphic to $Z_{pq}$ for distinct primes $p$ and $q$; this is basically Chinese Remainder Theorem. But this is not relevant here.) $\endgroup$ – quid May 17 '15 at 14:37
  • $\begingroup$ I did never say that $\alpha_i=1$. I said $n=1$ and in the examples I gave there is just no exponent for convenience of the presentation. $\endgroup$ – quid May 17 '15 at 14:48
  • $\begingroup$ Yes, prime power order. The order is a power of a prime. $\endgroup$ – quid May 17 '15 at 14:51
  • $\begingroup$ sorry i misunderstood that $\endgroup$ – Learnmore May 17 '15 at 14:52

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