# Finite Abelian group with the given property [duplicate]

Let $G$ be a finite Abelian group such that it contains a subgroup $H_0 \neq (e)$ which lies in every subgroup $H \neq (e)$. Prove that $G$ must be cyclic. What can you say about order of $G$ ?

I have examples to illustrate this result, but i can no idea about the general proof.

Here's my examples:

1) Take $G=\Bbb{Z}_8$ and $H_0=\{0,4\}$. Clearly $H_0$ lies every other subgroups $(\neq(e))$, namely lies in $\{0,2,4,6\}$ and lies in $\Bbb{Z}_8$

Here, $o(G)=8=2^3$

Edit:

2) Similarly, take $G=\Bbb{Z}_{16}$ and $H_0=\{0,8\}$. Clearly $H_0$ lies every other subgroups $(\neq(e))$, namely lies in $\{0,4,8,12\}$, $\{0,2,4,6,8,10,12,14\}$ and lies in $\Bbb{Z}_{16}$

Here, $o(G)=16=2^4$

3) Likewise, $G=\Bbb{Z}_{4}$ and $H_0=\{0,2\}$. Clearly $H_0$ lies every other subgroups $(\neq(e))$, namely lies in $\Bbb{Z}_{4}$

Here, $o(G)=4=2^2$

From this examples, can i conclude $o(G)=p^n$, for a prime $p$ ?

Because, if $o(G)=p^n$, then there are unique subgroups of order $1,p,p^2,p^3,..,p^{n-1}$ and in this case we take $H_0$ as a subgroup of order $p$, then it is immediately lies in other subgroups.

Is this correct ? What is the general proof?

• Suppose $G$ is cyclic of order $p$...then the only subgroups are $e,G$ so the property holds. What about order $p^2$? $p^4$? – lulu Jul 25 '17 at 11:40

By Cauchy's theorem, $G$ has subgroups of order $p$ for all primes $p$ that divide the order of $G$. Since these subgroups for different primes intersect only at the trivial subgroup, there can be only one prime divisor and so $G$ is a $p$-group.