A group $G$ is indecomposable if $G \neq \langle e \rangle$ and $G$ cannot be written as the direct product of two of its proper subgroups. Why is the additive group of rational numbers $(\mathbb{Q},+)$ indecomposable?
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$\begingroup$ Additional group?!! $\endgroup$– MikasaCommented Dec 29, 2013 at 14:17
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$\begingroup$ what do you mean by "decomposition"? $\endgroup$– user87543Commented Dec 29, 2013 at 14:17
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3 Answers
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I suspect, what you want to ask is why $(\mathbb{Q},+)$ is indecomposable, ie. it cannot be written as the direct sum of two subgroups.
The answer is that two non-trivial subgroups must intersect non-trivially. If $\{0\}\neq H, K < \mathbb{Q}$, then choose non-zero $p/q \in H, a/b\in K$, then $$ qa\frac{p}{q} = ap = pb\frac{a}{b} \in H\cap K\setminus \{0\} $$
answered Dec 29, 2013 at 14:23
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I assume you are asking why the additive group $Q$ of rationals is not a non-trivial direct sum.
If it were, then the projection on one of the direct summands would be a non-trivial idempotent in the endomorphism ring of $Q$. But the endomorphism ring of $Q$ is isomorphic to the ring of rationals. This is a field, so it has only the trivial idempotents 0, 1.
answered Dec 29, 2013 at 14:27
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Because there are no additive primes. Every number is the sum of two other nontrivial numbers.
answered Dec 29, 2013 at 14:20