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Let $H$ and $K$ be group and $H\rtimes_{\phi} K$ is the semidirect product of those group by the homomorphism $\phi:K\rightarrow Aut(H)$

Now , i have a main question about this function $\phi$

Does $\phi$ map every element $k\in K$ into the same element $\sigma$ in $Aut(H)$?

I mean that , is $ \forall k_1 ,k_2 \in K ,\phi (k_1)=\phi (k_2) =\sigma $ ?

in the examples of D&F , I have noticed that in the first example, $\phi$ maps every element of K to the automprphism of inversion ( the automorphism which map the element $x$ to $x^{-1}$ and the same thing in other examples ) , so , is this true every where ?

If this is true , then there is a problem which is , this doesn't give us a homomorphism , if $\phi$ maps the elements to the inversion automorphism , this $\phi$ is not a homomorphism as $\phi (x_1 x_2) = \tau$ but $\phi (x_1) \phi (x_2)= I$ where $\tau$ is the automorphism of inversion and $I$ is the identity automorphism

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  • $\begingroup$ What is D&F? What first example? $\endgroup$ – martini Jun 26 '13 at 11:32
  • $\begingroup$ @martini ,D&F is Dummit and Foote , the example prove that , $D_{2n} \cong H\rtimes_{\phi} K$ where $H=Z_n , K=Z_2$ and $\phi$ is the automorphism which maps every element to its inverse $\endgroup$ – Fawzy Hegab Jun 26 '13 at 11:41
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No, $\phi$ is just an arbitrary map into $\operatorname{Aut}(H)$ (sometimes called an action of $K$ on $H$). Sometimes, however, this is restrictive. For example, if $H\cong\mathbb{Z}$ then $\operatorname{Aut}(H)\cong C_2$ (I write $C_n$ for the cyclic group of order $n$) and so every map from $C_5$, say, to $C_2$ is the trivial map. So $C_5\rtimes H\cong C_5\times H$.

For an example of elements acting with different automorphisms, consider $K=C_2\times C_2 \times C_2=\langle a, b, c\rangle$ and $H=C_3\times C_3=\langle x, y\rangle$ and take the action to be $x^a=x$, $y^a=y$, $x^b=x^{-1}$, $y^b=y^{-1}$ and $x^c=y$, $y^c=x$. This action is associated to a semidirect product, but clearly $a$, $b$ and $c$ act on $H$ in different ways and so are being mapped to different automorphisms of $H$ by $\phi$.

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Your $\phi$ is (just as stated) an arbitrary group homomorphism from the group $K$ to the group $\operatorname{Aut}(H)$.

If $\phi$ maps every $k\in K$ to the same element $\sigma\in\operatorname{Aut}(H)$, then necessarily $\sigma=\operatorname{id}_H$ and the semidirect product is simply the direct product.

In particular, the map $H\to H$, $x\mapsto x^{-1}$ is not an automorphism of $H$, unless $H$ is abelian. And even then you cannot map every element to the inversion automorphism (unless all elements are their own inverses and this is in fact the identity).

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  • $\begingroup$ , this is cool , in the example , they prove that $D_{2n} \cong Z_n \rtimes_{\phi} Z_2$ let $H=Z_2 , K=Z_2$ and $\phi:K\rightarrow Aut(H)$ defined by mapping x to the automorphism of inversion on $H$ so that the associated action is $x . h = h^{-1}$ for all $h\in H$ ,so is this a typo in the text ( Dummit and foote , the examples of semidirect product )? of this doesn't mean that every element is mapped to the inversion auomorphism ? $\endgroup$ – Fawzy Hegab Jun 26 '13 at 12:20

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