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Let $A$ and $x$ be $n \times n$ and $n \times 1$ matrices, all entries real and strictly positive. Assume that $A^2 x = x$. Show that $A x = x$.

Suppose that $(\mathbb Q,+)$ be the additive rational group and $H$ a subgroup of it. Which of the following statements is not true?

(a) If $\mathbb Q/H\cong \mathbb Q$, then $H=0$.

(b) If $H\neq 0$, then every proper subgroup of $\mathbb Q/H$ is of finite order.

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(c) If $H\neq 0$, then every element of $\mathbb Q/H$ is of finite order.

(d) If $\mathbb Q/H$ is finite, then $H=\mathbb Q$.

I have tried to solve it. I am not able to draw conclusions. Please help me.

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The additive group $\,\Bbb Q\,$ is as abelian *divisible group*, meaning:

$$\forall\,g\in \Bbb Q\,\,\wedge\,\,\forall n\in\Bbb N:=\{1,2,…\}\,\,\,\exists\,x\in \Bbb Q\,\,\,s.t.\,\,\,g=nx$$

Some facts that’d be interesting you can prove about divisible groups: if $\,G\,$ is a divisible group then

1) Every homomprphic image of $\,G\,$ is divisible

2) The only finite group that is divisible is the trivial group $\,\{0\}\,$

3) The group $\,G\,$ has no maximal subgroups

Thus, almost automatically we already have that (b) isn’t true (and this already solves your problem), (d) is true.

Now, (c) follows from the following: if we take $\,\displaystyle{0\neq h=\frac{p}{q}\in H}\,$ then *for any* $\,\displaystyle{x=\frac{a}{b}\in\Bbb Q}\,$ :

$$bp(x+H)=bp\left(\frac{a}{b}+H\right)=pa+H=(qa)\frac{p}{q}+H=qa\left(\frac{p}{q}+H\right)=qaH=H\Longrightarrow$$

$$\operatorname{ord}(x+H)\leq bp$$

And now finally (a) follows also from the above.

Hint: It can be easily verified that $\frac{\mathbb Q}{\mathbb Z}\cong\bigoplus_{p\in P}\mathbb Z(p^{\infty})$. So, I think we can work on $H=\mathbb Z\leq\mathbb Q$ and probe the right option.

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