I know there are methods of showing when $H\rtimes_{\Psi_{1}}K$$\cong$$H\rtimes_{\Psi_{2}} K$.

However, what about semidirect products in which the H's differ.

Is it ever the case where $|H_{1}|=|H_{2}|$, $H_{1} \ncong H_{2}$ but we have that $H_{1}\rtimes_{\Psi_{1}}K$$\cong$$H_{2}\rtimes_{\Psi_{2}} K$.

  • $\begingroup$ Wouldn't that contradict the assumption $|H_{1}|=|H_{2}|$ $\endgroup$ – Matthew Cheung Dec 1 '16 at 17:54
  • $\begingroup$ Sorry yes. A genuine example is the dihedral group of order $8$, in which your can take $H_1$ cyclic of order $4$ and $H_2$ a Klein $4$-group. $\endgroup$ – Derek Holt Dec 1 '16 at 18:14
  • $\begingroup$ Since the dihedral group is of order 8, I believe our K would be the group of order 2 which is $\mathbb{Z}2$ But I believe that if we semidirect product H1 which is $\mathbb{Z4}$ with K, we get the dihedral group, but if we semidirect product H2 which is the Klein four group with K, we get the quaternion group. $\endgroup$ – Matthew Cheung Dec 2 '16 at 7:02
  • $\begingroup$ The quaternion group of order $8$ has a unique element of order $2$, and it does not have $K$ as a subgroup. I have given more details of the example in my answer below. $\endgroup$ – Derek Holt Dec 2 '16 at 10:28

An example is the dihedral group of order $8$. Its elements are the rotations $1,a,a^2,a^3$, and the reflections $b,ab,a^2b,a^3b$.

We can take $H_1=\{1,a,a^2,a^3\}$, $H_2=\{1,a^2,ab,a^3b\}$, with $K=\{1,b\}$. Then $H_1$ is cyclic and $H_2$ is a Klein $4$-group.


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