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Well there's this exercise:

Let $\mathbb{K}^\times$ be a commutative field.

Let $\phi: \mathbb{K}^\times \times SL_n (\mathbb{K}) \rightarrow GL_n(\mathbb{K})$ be an isomorphism (that I've proven), such that $\phi(z,M)=\begin{vmatrix} z&0\\ 0&I_{n-1}\\ \end{vmatrix} M $

What would be the induced function for the semi-direct product of $\mathbb{K}^\times \rtimes_{f} SL_n (\mathbb{K})$ ? (I don't understand what they mean by that, in french: "loi de produit semi-direct induite sur $\mathbb{K}^\times \times SL_n (\mathbb{K})$ par $\phi$")

Just for illustration:

Later on the exercise they ask me to show (using what we've shown) that $S_n \cong A_n \rtimes_f P$ where $|P|=2$ and $P<S_n$.

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  • $\begingroup$ As a set, the external semi-direct product is a Cartesian product and so isn't the induced function just itself? $\endgroup$
    – Oliver Jones
    Nov 10, 2019 at 2:32
  • $\begingroup$ Well I thought it should be something that defines the product: $(k_1,M_1) (k_2,M_2)=(k_1 \Phi_{M_1} (k_2),M_1M_2)$ Therefore: $ \Phi: SL_n(\mathbb{K}) \rightarrow Aut(\mathbb{K})$ such that $\Phi(M_1) = \Phi_{M_1}$, so I would be searching this $\Phi_{M_1}$ that is induced by the original function defined before $\endgroup$
    – Lucas Tonon
    Nov 10, 2019 at 11:00
  • $\begingroup$ Oh, I misunderstood what you meant by induced function. Here's a link to an answer to your question: math.stackexchange.com/a/670128/55622 $\endgroup$
    – Oliver Jones
    Nov 10, 2019 at 22:39
  • $\begingroup$ By the way, I think your map should be $\Phi : {\Bbb K}^\times \rightarrow Aut(SL_n({\Bbb K}))$. $\endgroup$
    – Oliver Jones
    Nov 10, 2019 at 22:43

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