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I am trying to understand wreath products but cannot find an example of one being computed in depth.

I know that $D_{4}$ is isomorphic to $\mathbb{Z}_{2} \wr \mathbb{Z}_{2}$, but I cannot seem to understand why this would be true from definitions I read in textbooks.

This wreath product is just the semidirect product of the Klein-4 group and $\mathbb{Z}_{2}$. How exactly would this semidirect product be computed, and how can I see that this is isomorphic to $D_{4}$?

Chris
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  • Perhaps this post is helpful. – Dietrich Burde Feb 15 '21 at 10:50
  • I found this helpful: https://www.youtube.com/watch?v=2Ik9m84AT2E&t=1110s – waveman Feb 15 '21 at 12:04
  • Since you refer to a disconnect with the textbook definition, it would be helpful to see which textbook definition you are using. Wreath products can be defined abstractly and very concretely, and it is hard to give a good answer withouty knowing the full context. – ahulpke Feb 15 '21 at 15:57

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The wreath product is defined in terms of a semi-direct product. There is the restricted and the unrestricted wreath product, depending as one takes a direct sum or product in the definition. These agree for finite products.

You need a group action. In this case we get $\Bbb Z_2$ acting on $\Bbb Z_2^2$ by shifts. Then it is the semi-direct product $\Bbb Z_2^2\rtimes\Bbb Z_2$. It turns out that this is indeed the usual $D_4\cong\Bbb Z_4\rtimes\Bbb Z_2$. (Notice that the actions in these two semi-direct products are different.)

$\Bbb Z_2\wr\Bbb Z_2$ is the smallest non-trivial wreath product. It also equals the hyperoctahedral group $S_2\wr S_2$.

To see the isomorphism, you could work through the notation; or you can convince yourself that this is not the quaternions, the only other nonabelian group of order $8$.