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Is there a simple example which verify the following assertion : $ G / H_1 \cong G / H_2 $ and $ | G / H_1 | = | G / H_2 | = 2 $ and $ H_1 \neq H_2 $ ? $ G $ is a groupe. $ H_1 $ and $ H_2 $ are two subgroups normals of $ G $. Thanks to all people here in this web site.

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3 Answers

Let $G=\Bbb Z\times\Bbb Z$, $H_1=2\Bbb Z\times\Bbb Z$, and $H_2=\Bbb Z\times 2\Bbb Z$.

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Side question: What if equal were changed to isomorphic. –  Amr Dec 15 '12 at 19:56
    
Thank you very much. :) –  Bryan Dec 15 '12 at 19:58
    
@Amr: I don’t know, and I’m not nearly enough of an algebraist to trust my intuition. –  Brian M. Scott Dec 15 '12 at 19:59
    
@Bryan: You’re welcome. –  Brian M. Scott Dec 15 '12 at 19:59
    
@Amr: If you see my solution, there is such an example. –  Clayton Dec 15 '12 at 20:04
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Let $G=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$, $H_1=\langle(1,0),(0,2)\rangle$, and $H_2=\langle(0,1)\rangle$. Then $H_1\cong \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, while $H_2\cong \mathbb{Z}/4\mathbb{Z}$, but $|G/H_1|=|G/H_2|=2$.

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Thank you very much :) –  Bryan Dec 15 '12 at 19:59
    
@Clayton : I did the mistake before too, but $\Bbb Z_2$ is standard for $p$-adics ring $\Bbb Z_p$. You should use the notation $\Bbb Z/n \Bbb Z$ for the quotient of $\Bbb Z$. –  Patrick Da Silva Dec 15 '12 at 20:01
    
I normally do, but I was feeling lazy. I'll edit it. –  Clayton Dec 15 '12 at 20:03
    
That looks familiar ... –  Henning Makholm Dec 15 '12 at 20:50
    
@Henning Makholm: Haha, I didn't realize you had posted the same thing elsewhere. What a great example! –  Clayton Dec 15 '12 at 20:53
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Well, to begin with, if $|G/H_1|=|G/H_2|=2$, it follows immediately that $G/H_1\cong G/H_2$. There is only one group of order $2$.

An easy example is $D_4$. $D_4/\langle r \rangle$, where $r$ is the rotation of order $4$, is of order $2$, as is $D_4/\langle r^2s \rangle$, where $s$ is the transposition.

The way I think about finding these types of counterexample is looking for elementary abelian subgroups in nonabelian groups. In the Klein $V$ group, for example, $\mathbb{Z}_2$ occurs as $3$ different subgroups: $\langle (0,1) \rangle, \langle (1,1) \rangle,$ and $\langle (1,0) \rangle$. If you can find a Klein $V$ group in a nonabelian group, as I have above (here $r^2s$ corresponds to $(1,1)$), you can find other subgroups which intersect with the $V$ group $(1,0)$ part, such as the $\langle r \rangle$ group, and study how their interaction with the $(1,1)$ case.

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Thank you very much. :) –  Bryan Dec 15 '12 at 19:59
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