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Only when H$\rtimes$K is direct product and H,K are abelian? How to prove? In other words if the homomorphic from K to Aut(H) is not trivial then the semi direct product is not abelian?

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The comutativity is equivalent to

$$y_1 y_2= y_2 y_1 \forall y_i\in K$$ $$x_1 \phi ( y_1 ) ( x_2 ) = x_2\phi ( y_2 ) ( x_1 ) \forall x_i \in H y_i \in K$$

The $K$ must be abelian, from the second equation we take

$$\phi (y_2 ) ( x_1 ) = x_2^{-1} x_1 \phi ( y_1 ) ( x_2 ) $$

So $\phi (y_2 ) ( x_1 ) $ does not depend on the value of $y_2$, hence $\phi$ is a constant homomorphism, hence is the null homomorphism ( always equal to the identity ) then from the second equation again

$$x_2 x_1 = x_1 x_2 $$

So $H$ must be abelian as well. So the semi-direct product is abelian iff both factors are abelian and the linking homomorphism is the trivial.

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  • $\begingroup$ Sorry, I don't think it's clear that ϕ(y2)(x1) does not depend on the value of y2, because if change y2 to y3, then ϕ(y3)(x1)=x^(−1)3x1ϕ(y1)(x3) $\endgroup$ – 6666 Oct 1 '14 at 18:42
  • $\begingroup$ $x_3$ does not depend on $y_3$. $\endgroup$ – PenasRaul Oct 1 '14 at 20:49
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If $\phi:K\to{\rm Aut}(H)$ is not trivial then $\phi_k:H\to H$ is nontrivial for some $k\in K$, so $\phi_k(h)\ne h$ for some $h\in H$, in which case $khk^{-1}=\phi_k(h)\ne h~\,\Rightarrow~ kh\ne hk$ within $H\rtimes K$.

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$H$ and $K$ are subgroups of $H\rtimes K$, so if $H\rtimes K$ then $H$ and $K$ had better be abelian too.

The homomorphism $\varphi:K\to\operatorname{Aut}(H)$ becomes conjugation in $H\rtimes_\varphi K$. Conjugation is always trivial in abelian groups, so again, if $H\rtimes_\varphi K$ is abelian, then $\varphi$ had better be trivial.

In fact, the following statement is true: $H\rtimes_\varphi K$ is abelian if and only if $H$ is abelian, $K$ is abelian, and $\varphi$ is the trivial homomorphism.

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