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The following question is for my own curiosity as I take some time to get reacquainted with group theory.

Let G be a semi-direct product of the groups N and K with multiplication defined by the automorphism $\phi$ from K to Aut(N). Let Fix($\phi$) be the set of all elements of N that are mapped to themselves by all elements of the range of $\phi$. Clearly every element of Fix($\phi$) commutes with all elements of K and every element of the kernel of $\phi$ commutes with every element of N.

If Fix($\phi$) is the trivial group in N and Ker($\phi$) is the trivial group in K, does that imply that the center of G is trivial? If so, could someone point me to a reference or proof. If not, then a counter example.

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    $\begingroup$ The image of the center under a surjective homomorphism is contained in the center of the image. Apply this to $G \to K$. $\endgroup$
    – Someone
    Commented Nov 29, 2010 at 17:41
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    $\begingroup$ Let me add just a tiny remark: if Fix$(\phi)$ is not trivial, then the centre of the group might be bigger than meets the eye at first glance: you can have $K$ acting faithfully (i.e. Ker$(\phi) = 1$), but through an inner automorphism. Then, you might be able to replace $K$ by a different complement such that suddenly $G$ falls apart into a direct product. As a concrete example, consider $N=D_4$, the dihedral group with 8 elements, and let $K$ act on it through the inner automorphism of order 4. The centre will be considerably bigger than what you might first think. $\endgroup$
    – Alex B.
    Commented Nov 30, 2010 at 1:45
  • $\begingroup$ @Alex: Which inner automorphism of $D_4$ has order 4? $D_4/\mathrm{Z}(D_4)$ is elementary abelian of order 4. $\endgroup$
    – Someone
    Commented Nov 30, 2010 at 9:02
  • $\begingroup$ @Someone I am sorry, you are right, $D_4$ was a bad example. Take $D_8$ instead. $\endgroup$
    – Alex B.
    Commented Dec 4, 2010 at 9:20

1 Answer 1

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Suppose that $z=xy$ is in the centre where $x\in N$ and $y\in K$. Then for all $u\in K$, $uxy=xyu$. But $uxy=\phi(u)(x)uy$ so that $x=\phi(u)(x)$ (and $uy=yu$). As this is true for all $u\in K$ then by the assumption on Fix($\phi$), $x=1$. Therefore $z=y\in K$.

As $y$ commutes with all elements of $N$ then $y$ lies in Ker($\phi$) and is trivial. So $z=1$ and the centre of $G$ is trivial.

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  • $\begingroup$ @Robin Chapman. Thanks. I was on the right track. I didn't stop to consider how each element of G factoring uniquely into two elements, one from N and one from K, would imply uy = yu and $\phi$(u)(x) = x. $\endgroup$
    – Greg Gibson
    Commented Nov 29, 2010 at 20:52

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