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.
An example of groups that $ G / H_1 \cong G / H_2 $ and $ | G / H_1 | = | G / H_2 | = 2 $ and $ H_1 \neq H_2 $
3 Answers
Let $G=\Bbb Z\times\Bbb Z$, $H_1=2\Bbb Z\times\Bbb Z$, and $H_2=\Bbb Z\times 2\Bbb Z$.
-
0@Amr: If you see my solution, there is such an example. – 2012-12-15
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 $|G/H_1|=|G/H_2|=2$.
-
0@Henning Makholm: Haha, I didn't realize you had posted the same thing elsewhere. What a great example! – 2012-12-15
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.