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I'm trying to find the minimal generating set for a square prism under reflection. This group is $\text{Dih}_4\times Z_2$.

Geometrically, the set $\{a,b,c\}$ where $a$ is a rotation of $90^\circ$ about the axis through the square faces, $b$ is a rotation of $180^\circ$ about an axis through two rectangular faces, and $c$ is a reflection about some plane, would generate the group. After all, $\{a,b\}$ generates $\text{Dih}_4$ and $\{c\}$ generates $Z_2$. However, I don't know how to prove that there does not exist two elements $\{x,y\}$ which generate the whole group.

Is this even the case? In any case, how can it be proven or disproven?

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    $\begingroup$ Can you show that the group you have has a quotient that is isomorphic to $Z_2\times Z_2\times Z_2$? $\endgroup$ – Arturo Magidin Jun 5 at 1:48
  • $\begingroup$ I think so; is it sufficient to show that $Z_2\times Z_2$ is a quotient of $\text{Dih}_4$? $\endgroup$ – Ovinus Real Jun 5 at 2:16
  • $\begingroup$ Yes, since then you get that $Z_2\times Z_2\times Z_2$ is a quotient of $D_4\times Z_2$. So, given that, suppose you had a $2$-element generating set for $D_4\times Z_2$. Would the images of the elements generate any quotient? If so, can $Z_2\times Z_2\times Z_2$ be generated by $2$ elements? $\endgroup$ – Arturo Magidin Jun 5 at 2:43
  • $\begingroup$ I see, so the logic is if some set $S\subseteq D_4\times Z_2=G$ can generate $G$, and there is a normal subgroup $H$ such that $|G/H|=Z_2\times Z_2\times Z_2$, then the cosets of $H$ with respect to $S$ also generate $G/H$, so since the minimal generating set of $Z_2\times Z_2\times Z_2$ clearly has cardinality $3$, we have $|\text{cosets}|\geq 3\Longrightarrow |S|\geq 3$? $\endgroup$ – Ovinus Real Jun 5 at 3:10
  • $\begingroup$ It’s not that the number of cosets is at least $3$ (that assertion is unclear). It’s that any generating set for $G$ maps to a generating set for $G/N$ for any normal subgroup $N$ of $G$. So if you know that there is an such that $G/N$ needs at least $k$ generators, then $G$ needs at least $k$ generators as well (though it may need more). $\endgroup$ – Arturo Magidin Jun 5 at 3:23

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