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!

  • 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

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.

  • 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

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.

  • $\begingroup$ 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

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