13
$\begingroup$

I know this has been asked before, in a bunch of places, so I expect it to be closed as a duplicate at some point, but I'd like to know if my own proof works. If the proof is incorrect, please limit answers to pointing out the errors. If the proof is correct, then thoughts on improving it would be appreciated.

Prove that $\Bbb Q$ is not the direct product of two non-trivial groups.

—Aluffi exercise II.3.5.

My proof

Suppose that $\Bbb Q = G \times H$, neither $G$ nor $H$ trivial.

First note that $0_G$ is the only element of $G$ with finite order: If $m\ne 0$ and $m\cdot g=0_G$, then $m\cdot(g,0_H)=0_{G\times H}$, contradicting the fact that $\Bbb Q$ has no non-zero elements of finite order.

Suppose that for some $m\ne 0$ and some $n\ne 0$, $\pi_G(m/n)=0_G$. Then $0_G=n\cdot\pi_G\left(\frac m n\right)=\pi_G\left(n\cdot\frac m n\right)=\pi_G(m)=m\cdot\pi_G(1)$.

Since $\pi_G(1)$ has finite order, we conclude that $\pi_G(1)=0$. Since $\Bbb Z=\langle 1\rangle$, $\pi_G$ maps every integer to $0_G$.

Now let $a$ and $b$ be non-zero integers. Then $b\cdot\pi_G\left(\frac a b\right)=\pi_G\left(b\cdot \frac a b\right)=\pi_G(a)=0_G$. As above, this shows that $\pi_G\left(\frac a b\right)=0_G$.

Thus if $\pi_G$ is not an injective function, then it must map everything to $0_G$, so $G$ will be a trivial group. The same result holds for $\pi_H$. Since by assumption, $G$ and $H$ are non-trivial, $\pi_G$ and $\pi_H$ are both injective functions, which is of course impossible.

$\endgroup$
3
  • $\begingroup$ I'm sorry, but I don't understand when you say that $0_G$ is the only non-zero element of $G$ with finite order. Isn't $0_G$ the zero element of $G$? $\endgroup$
    – tylerc0816
    Dec 7, 2013 at 1:32
  • 1
    $\begingroup$ @tylerc0816, fixed. $\endgroup$
    – dfeuer
    Dec 7, 2013 at 1:46
  • $\begingroup$ Looks like a good argument to me... $\endgroup$
    – Igor Rivin
    Dec 7, 2013 at 2:08

3 Answers 3

7
$\begingroup$

Yes, your proof is correct. You can avoid proof by contradiction though (which IMO is a good thing to do): write $\mathbb{Q} = G \times H$. WLOG $G$ is nontrivial. Then your argument shows that $\pi_G$ is injective, but its kernel is $0 \times H$, therefore $H$ is trivial. QED.

Is there a specific part of your proof that made you doubt whether it was correct?

$\endgroup$
4
  • $\begingroup$ I always doubt myself when I'm dealing with anything to do with algebra. I'm still trying to understand your answer, but I'm having trouble matching up the pieces with mine. Do you think you could add a bit more context around the bits? $\endgroup$
    – dfeuer
    Dec 7, 2013 at 3:36
  • $\begingroup$ @dfeuer: the proof is almost the same. The only thing that differs is that you assume that $H$ is non-trivial and then show that it must be trivial to derive a contradiction. But assuming that $H$ is non-trivial wasn't necessary anywhere in your proof, so you can just remove that assumption. It's enough that $G$ is non-trivial because then $\pi_G$ is injective hence its kernel (which is obviously equal to $0 \times H \cong H$) is trivial, so that $H$ is trivial. $\endgroup$
    – Marek
    Dec 7, 2013 at 16:40
  • $\begingroup$ @Marek: I'm not familiar with the convention in group theory, but when you two say that $\ker\pi_G$ is $0\times H$, do you mean $\{e_G\} \times H$? And now since $\pi_G : G\times H \to G$ is an injection, then $\ker\pi_G = \{(e_G,e_H)\}$, and so $H$ has to be $\{e_H\}$, right? The bulk of the proof seems to be proving that the natural surjection $\pi_G$ is also an injection. $\endgroup$ Apr 5, 2015 at 5:41
  • 1
    $\begingroup$ Right on both accounts @Andrey. $\endgroup$ Apr 5, 2015 at 8:53
3
$\begingroup$

HINT:

Any two non-zero subgroups have a non-zero intersection, since any two non-zero elements have a common multiple.

$\endgroup$
-1
$\begingroup$

Here's my try:

Suppose $\mathbb{Q} \cong H \times G$, where $G$ is non trivial. If there is nonzero $g \in G$ of finite order $n$, then $(e_H, g)^n = (e_H, e_G)$, which can't hold because of the isomorphism.

So, $\vert G \vert \geq \vert \mathbb{N} \vert$, and $\vert G \vert \leq \vert \mathbb{Q} \vert = \vert \mathbb{N}\vert$. (Because otherwise $\vert G \times H \vert > \vert \mathbb{Q} \vert$)

$\Longrightarrow \vert G \vert = \vert \mathbb{Q}\vert$, now we can say that

considering $f:\mathbb{Q} \longrightarrow H \times G $, $\forall (e_H, g) \in G \times H$ $\exists q \in \mathbb{Q} :f(q) = (e_H, g) \Longrightarrow \mathbb{Q} \cong \{e\} \times G$.

$\endgroup$
3
  • $\begingroup$ If you are wanting someone to check your proof then you should as this as anew question. $\endgroup$
    – user1729
    Mar 1, 2018 at 17:12
  • 1
    $\begingroup$ If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review $\endgroup$ Mar 1, 2018 at 17:21
  • $\begingroup$ @Taroccoesbrocco: Thank you, I will keep this in mind next time. I already realized that it is wrong, since the fact that $\vert G \vert = \vert \mathbb{Q} \vert$ In general doesn't imply that every $q \in \mathbb{Q}$ has it's image in $\{e\} \times G$. Counterexample: pick $\mathbb{Z} \times \mathbb{Z}$ instead of $\mathbb{Q}$ and $H = G = \mathbb{Z}$. $\endgroup$ Mar 2, 2018 at 11:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .