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In a certain exercise I am asked to prove something that involves semidirect product. I wanted to know if $\mathbb{Z}$ could be embedded in a semidirect product of bigger groups, named for example $A,B$, this is, if there exists, $A,B$ so that $\mathbb{Z} < A\rtimes_{\theta} B$ for some $\theta: B \rightarrow Aut(A)$. Is it possible to simply assume this?

Emmy N.
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    You can clearly do this for any group $A$ or any group $B$ that has ${\mathbb Z}$ as a subgroup (or equivalently has an element of infinite order). – Derek Holt Nov 01 '23 at 22:34
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    ... or when in the new group $G=\mathbb{Z} < A\rtimes_{\theta} B$ there is an element of infinite order $g$, the injective group map being $n\to g^n$. – dan_fulea Nov 01 '23 at 22:39

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Any direct product is a semidirect product, so just take

$$\Bbb Z(\cong\circ \le)\Bbb Z\times \Bbb Z_2,$$

where $(\cong\circ \le)$ means "isomorphic to a subgroup of".

Less glib, look at the semidirect products of $\Bbb Z$ with itself, discussed here.

Shaun
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