Does the group $(\mathbb{R},+)/(\mathbb{Q},+)$ have a subgroup of order 5?

Question

Does the group $(\mathbb{R},+)/(\mathbb{Q},+)$ have a subgroup of order 5?

I'm really struggling with how to answer this question and would appreciate any help. Thank you so much!

Answer 1

Every group of order $p$ is cyclic. If $\mathbb{R}/\mathbb{Q}$ were to have a subgroup $H$ of order $5$, then $H = \langle x + \mathbb{Q} \, |\, x \in \mathbb{R} \rangle$ such that $5 (x + \mathbb{Q}) = 5x + \mathbb{Q} = \mathbb{Q}$. This means that $5x \in \mathbb{Q}$, so that $5x = \frac p q$ or $x = \frac {p}{5q}$. But then this means that $H = \mathbb{Q}$ is the trivial subgroup, a contradiction.

Answer 2

Hint: In fact, $(\mathbb{R},+)/(\mathbb{Q},+)$ contains no nontrivial elements of finite order.

Indeed, $n \bar\alpha = 0$ iff $n \alpha \in \mathbb Q$ iff $\alpha \in \mathbb Q$ iff $\bar\alpha = 0$.

Answer 3

Suppose there is $x\in\Bbb R$ with $5[x] = [0]$. This means $5x\in\Bbb Q$ and then...

Answer 4

In questions like this, the first thing to look at is definitions, definitions, and definitions. What does "element of order 5" actually mean in this case? It means an element $x$ such that $$ x\neq 0\\x+x\neq 0\\x+x+x\neq 0\\x+x+x+x\neq 0 $$ but $x+x+x+x+x = 0$

And what is this group? It's a quotient group of the real numbers. So each element is of the form $r + \Bbb Q$ for some $r\in \Bbb R$ (I will write this element as $\bar r$), and addition goes by $\bar r + \bar s = \overline{r+s}$. Finally, $\bar r = \bar 0$ if and only if $r$ is rational.

So, what the question is actually asking you about, is whether there is a real number $r$, such that $$r\\ r+r\\ r+r+r\\ r+r+r+r$$ are all irrational, but $r+r+r+r+r$ is rational?

Answer 5

Both $\Bbb R$ and $\Bbb Q$ are $\Bbb Q$-vector spaces, and so is the quotient. which is thus torsion-free.