I'm reading "Contemporary Abstract Algebra," by Gallian.
This is (part of) Exercise 26 of the supplementary exercises for chapters 1-4 ibid., although I am requesting a proof of the contrapositive just out of interest, preferably using prior material available from the textbook.
I intend to use abstract algebra, keeping in the spirit of the textbook; thus proofs that rely on, say, real analysis (too much), will not be accepted.
$\mathcal{O}$: Suppose that $H$ and $K$ are nontrivial subgroups of $\Bbb Q$ under addition. Show that $H \cap K$ is a nontrivial subgroup of $\Bbb Q$.
The contrapositive problem can be stated as follows:
$\mathcal{C}$: Let $H$ and $K$ be subgroups of $\Bbb Q$. Suppose $H\cap K=\{0\}$ is trivial. Show that either $H$ or $K$ is trivial.
Thoughts:
To me, it seems $\mathcal{C}$ is one of those things that are so tauntingly "obvious", it's hard to know where to begin.
My problem is also that my only thought I have worth sharing for this is to try assuming - no joke! - that neither $H$ nor $K$ is trivial, then deducing $H\cap K$ is nontrivial, contradicting the hypothesis that $H\cap K=\{0\}$. That's just silly.
A lovely proof of the original statement $\mathcal{O}$ is given here, with variations on its theme here and here.
Please help :)
Let $P$ denote the property that triviality of an intersection of two subgroups implies triviality of either.
1: If all $G_i$ have the property $P$, then so does
$$\lim_{\longrightarrow}G_i .$$
2: Define $\mathbb{Z}^{(n)}=\mathbb{Z}$, and for $n|m$,
\begin{align} f_{n,m}:\mathbb{Z}^{(n)} &\longrightarrow\mathbb{Z}^{(m)}, \\ k &\mapsto\frac{m}{n}\cdot k. \end{align}
Then $$\mathbb{Q}=\lim_{\longrightarrow}\mathbb{Z}^{(n)}.$$
3: $\mathbb{Z}$ has the property $P$. In fact, $H\cap K=0$ implies the injectivity of $$\mathbb{Z}\longrightarrow \mathbb{Z}/H \times \mathbb{Z}/K,$$ and thus $H=0$ or $K=0$.