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Let $D_8 := \langle a,b \mid a^4 = 1 = b^2, bab = a^{-1}\rangle$

I'm trying to formally show that

$$D_{8} \cong C_4 \rtimes C_2 = \langle s\rangle \rtimes \langle t \rangle$$

My book gives as hint to consider the map $\phi: C_2 = \langle t \rangle \to Aut(C_4) = \{id, .^{-1}\}\cong C_2 $ that sends $1 \mapsto id, t \mapsto .^{-1}$ and then show that

$$D_8 \cong C_4 \rtimes_\phi C_2$$

The right-action associated with this map is given by:

$$*: C_4 \times C_2 \to C_4:$$

where $s*t =(s)((t)\phi))$ (I follow the convention in Isaac's algebra book where I write $(x)f$ instead of $f(x)$).

I tried to define an isomorphism explicitly:

$$D_8 \to C_4 \rtimes_\phi C_2$$

by mapping

$$a^jb^k \mapsto (k,j) \in \mathbb{Z}_2 \times \mathbb{Z}_4 \cong C_2 \times C_4$$

where $j \in \{0, \dots, 3\}, k = 0,1$.

This seems to be the only sensible choice for such a map.

How can I show that this map is a bijection (I see that injection or surjection is sufficient, and I also know that this is well-defined), and thus an isomorphism? Is there an easier way to see the isomorphism?

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  • $\begingroup$ Which book are you using? $\endgroup$ – Shaun Apr 1 at 10:36
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    $\begingroup$ Martin Isaacs "Algebra, A graduate course". $\endgroup$ – user370967 Apr 1 at 10:38
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    $\begingroup$ Since $b^2=e$, $b^{-1}=b$, so the LHS of relation $bab=a^{-1}$ can be written as conjunction of $a$ by $b$, so one has an inner automorphism . . . $\endgroup$ – Shaun Apr 1 at 10:42
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    $\begingroup$ I'm not sure if I can. My understanding of semidirect products is based on their presentations. However, here is a question that'll be of some help, I reckon :) $\endgroup$ – Shaun Apr 1 at 10:48
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    $\begingroup$ Thank you! I will look into it! $\endgroup$ – user370967 Apr 1 at 10:51

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