2
$\begingroup$

I was making some exercises on group theory when I came across this problem: Suppose G is a finite group and K and H are 2 different subgroups of index 2. Now first prove $(H \cap K) \triangleleft G$ and then prove $\frac{G}{H \cap K}$ is not cyclic. The first part went well (I used the fact that $H \triangleleft G$ and $K \triangleleft G$ because the indices of these subgroups are 2, the rest is quite easy). The second part however, didn't.

Thank you for your help!

$\endgroup$
  • 6
    $\begingroup$ How many subgroups of index 2 does a cyclic group have? $\endgroup$ – Jack Schmidt Aug 5 '13 at 19:03
  • $\begingroup$ Does this mean $\frac{G}{H\cap K}$ is $\textit{never}$ cyclic. With no other condition this isn't true because you can consider $H=K=\langle e\rangle\subset\mathbb{Z}/2\mathbb{Z}$. If it means is not necessarily cyclic then consider the example, $H=\{e\}\times\mathbb{Z}/2\mathbb{Z}, K=\mathbb{Z}/2\mathbb{Z}\times\{e\}\subset \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}=G$. $\endgroup$ – Owen Sizemore Aug 5 '13 at 19:22
  • 4
    $\begingroup$ You also need $H\neq K$ for this. $\endgroup$ – Tobias Kildetoft Aug 5 '13 at 19:26
  • $\begingroup$ Sorry, I had to translate the question from dutch... It says you may also assume that $H \neq K$. $\endgroup$ – Leo Aug 5 '13 at 20:35
  • $\begingroup$ I'll correct it then... $\endgroup$ – Leo Aug 5 '13 at 21:03
4
$\begingroup$

This is Jack's comment (hint) fleshed out: suppose $\,G/(H\cap K)\;$ is cyclic and $\,H\ne K\;$ :

$$H/(H\cap K)\;,\;K/(H\cap K)\le G/(H\cap K)\;,\;\;\text{and we also have that}$$

$$\left|H/(H\cap K)\right|\cong \left|HK/K\right|\stackrel{\text{why?}}=2\stackrel{\text{why?}}=\left|HK/H\right|\cong\left|K/(H\cap K)\right|$$

Thus we have a finite cyclic group with two subgroups of the same order, and this is impossible.

$\endgroup$
  • 1
    $\begingroup$ Ok, I think I understand. The answer to your educational question is $HK = G$, isn't it? $\endgroup$ – Leo Aug 5 '13 at 21:02
  • $\begingroup$ Bingo, you got it. $\endgroup$ – RghtHndSd Aug 6 '13 at 1:58
3
$\begingroup$

As mentioned in the comments, we need $H \neq K$. Let $\alpha \in G$. Then it is easy to see that $\alpha^2 \in H$ and $\alpha^2 \in K$ as $[G : H] = [G : K] = 2$. Thus, $\bar{\alpha}^2 = \bar{1}$. If $\bar{\alpha}$ generates $G/(H \cap K)$, then it must be that $[G : H \cap K] = 2$ which is not possible.

$\endgroup$
  • $\begingroup$ I don't see why $\alpha^2 \in K$ or why $\alpha^{2} \in H$. Obviously this will be the same reason. Could you please elaborate? Thanks! $\endgroup$ – Leo Aug 5 '13 at 20:58
  • $\begingroup$ With $\alpha \in G$, consider $\alpha H \in G/H$. Remembering that $G/H$ has order 2, we must have $(\alpha H)^2 = H$, or rather $\alpha^2 H = H$ making $\alpha^2 \in H$. $\endgroup$ – RghtHndSd Aug 5 '13 at 21:17
  • $\begingroup$ Ok, thank you... It seems I still have a lot to learn... $\endgroup$ – Leo Aug 5 '13 at 21:29
  • $\begingroup$ Do you expect there will be a time when that won't be true? Because there won't. $\endgroup$ – RghtHndSd Aug 5 '13 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.