# Finite abelian group with common nontrivial subgroup $H_0$ is cyclic.

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 $G$ is cyclic. what is order of $G$? i try to solve this problem, but i can not finish..kindly help me, or give any other solution. I try, since $G$ is finite abelian group, so it isomorphic direct product of its sylow subgroup, $G\thickapprox S(P_{1})*S(P_{2}*)....S(P_{k})$ since $H_{0}\subset S(P_{i})$ and $S(P_{i})\cap S(P_{j})= \phi$ this imlies $k=1$ i.e. $G\thickapprox S(P_{1})$, from here how i conclude $G$ is cyclic?

Actually the intersection of two distinct Sylow subgroups is $\{e\}$ rather than the empty set. A finite abelian group is a product of cyclic groups. If there is more than one group in the product, you again get a pair of subgroups with trivial intersection $\{e\}$, and therefore they couldn't both contain $H$, giving the required contradiction.

Here is a proof that only uses the Sylow decomposition, not the decomposition into cyclic groups. Your argument reduces us to the case that $G$ has order $p^n$ for some prime $p$. Let $x$ be an element of maximal order, say of order $p^n$. Then $H_0$ must be the group generated by $p^{n-1}x$, since there can only be one order $p$ subgroup. Now let $y$ be some element not in the subgroup generated by $x$ of minimal order, say $p^m$. Then there must be some $0 < k < p$ with $p^{m-1}y = kp^{n-1}x$. Then $p^{m-1}(y - kp^{n-m}x) = p^{m-1}y - kp^{m-1}x = 0$. Since the order of $y - kp^{n-m}x$ is smaller than $m$, minimality gives $y - kp^{n-m}x \in \langle x \rangle$. So $y \in \langle x \rangle$. So $G$ is cyclic.

• This is really nice! – user276011 Jul 15 '19 at 1:41

Let's start with $$o(G) = p^n$$. From Sylow theorem, there exist an element $$x \in G$$ of order $$p$$. Thus $$H_0 = \{x^i|i\in[1,p]\}$$.

Now we will prove that, if for each two cyclic subgroups $$H_a,H_b$$ of orders $$p^s,p^t$$ (with $$s\geq t$$) generated by $$a,b$$ respectively we have $$o(H_a\cap H_b) \geq p^{t-1}$$,

then, $$H_b \subset H_a$$

\begin{align*} a^{p^{s-1}}&=b^{p^{t-1}} \\ (a^{p^{s-t}}b^{-1})^{p^{t-1}}&=e \\ a^{p^{s-t}}b^{-1} &= b^p (\text{or any } b^{kp}, k We know that $$o(b^{p+1}) = p^t$$

Hence, $$H_b \subset H_a$$

When you apply induction using above lemma, we see that $$G$$ is cyclic.