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I have been asked to:

Decide whether the following groups are decomposable:

(a) - $(\mathbb{R^*}, \cdot)$

(b) - $(\mathbb{C}, +)$

(c) - $(\mathbb{Q^*}, \cdot)$

(d) - $(\mathbb{Q}, +)$

I would like a hint for item (a). I believe I was able to do itens (b), (c) and (d).

Regarding item (a), I tried to decompose $\mathbb{R^*}$ in rationals and irrationals (but this failed, since the irrationals are not a subgroup) or into algebraic and transcendental numbers (which also fails, since the transcendental numbers are not a subgroup). I also thought about showing that if $\mathbb{R^*} = A \times B$ then $A$ and $B$ do not intersect trivially (thus showing that the group is indecomposable), but I couldn't prove this idea.

Regarding item (b), I decomposed $\mathbb{C}$ into $\mathbb{R}$ and $i\mathbb{R} = \{iy \ | \ y \in \mathbb{R} \} $.

Regarding item (c), I wrote that $\mathbb{Q^*} = \langle \ p \ | \ p \ \text{is a prime} \rangle = \langle 2 \rangle \ \oplus \ \langle \ p \ | \ p \ \text{is an odd prime} \rangle $.

EDIT: As pointed in the comments, this decomposition is for the multiplicative group of positive rational numbers. A correct decomposition would be, for instance, $\mathbb{Q^*} = \langle 2, -1 \rangle \ \oplus \ \langle \ p \ | \ p \ \text{is an odd prime} \rangle $.

Regarding item (d), I proved that the group is indecomposable by proving that two non-trivial subgroups don't intersect trivially. My reasoning was the same as in: Why is the additive group of rational numbers indecomposable?.

Can anyone give me a hint for item (a)? Thanks in advance.

Community
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Gabriel F. Silva
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    Hint: Think of signs! – Aphelli Nov 11 '19 at 19:21
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    Also, note that (c) is one of many decompositions; in general, you can partition the primes into two sets and write $\mathbb{Q}^*$ as the direct sum of the subgroup generated by the primes in the first set and the subgroup generated by the primes in the second... – Steven Stadnicki Nov 11 '19 at 19:29

2 Answers2

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For item (a), you can decompose where one factor is the sign and the other is the absolute value. Notice that you probably should do (c) the same way. Your decomposition is for the multiplicative group of positive rational numbers.

It turns out $(\mathbb R, +) $ is decomposable as well, but it's not quite as easy to see this.

Matt Samuel
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    Good catch on (c), though OP's decomposition of $\langle \mathbb{Q}^*, \cdot\rangle$ seems relatively fine as long as $-1$ is treated as a prime and assigned to one subgroup or the other. – Steven Stadnicki Nov 11 '19 at 20:39
  • Thanks. And thank you also for noticing my mistake in item (c). I have edited the question to present a correct decomposition for $\mathbb{Q^*}$. – Gabriel F. Silva Nov 12 '19 at 01:10
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$(\mathbb R^*, \cdot) = \left(\mathbb R^+, \cdot\right)\times \{1, -1\}$, because (i) $\left(\mathbb R^+, \cdot\right)$ and $\{1, -1\}$ are clearly normal subgroups of $(\mathbb R^*, \cdot)$, (ii) common element of $\left(\mathbb R^+ , \cdot\right)$ and $\{1, -1\}$ is the identity $1$ only, and clearly (iii) $\mathbb R^* = \left(\mathbb R^+\right)\{1, -1\}$. Here $\mathbb R^*$ is the set of all non-zero real numbers and $\mathbb R^+$ is the set of all positive real numbers.

Mohsen Shahriari
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J. B.
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