1
$\begingroup$

Let $A$ be a finite abelian group and $G$ a finite non-abelian groups and let $\alpha$, $\beta:G\rightarrow{\rm Aut}(A)$ two homomorphisms with the same non-trivical kernel $H$.

Suppose that $\alpha (G)$ and $\beta (G)$ are conjugate subgroups of ${\rm Aut}(A)$. Are the semidirect products $A\rtimes _{\alpha }G$ and $A\rtimes _{\beta }G$ isomorphic?

I know that this is not true in general, but I can't find a counterexample.

Thank you in advance.

$\endgroup$
3
  • 3
    $\begingroup$ You are asking too many questions, and you have answered the first one yourself. Yes your example is correct. Of course there are other examples. $\endgroup$
    – Derek Holt
    Sep 2, 2022 at 17:07
  • 1
    $\begingroup$ And please don't ask new questions in comments! $\endgroup$
    – Derek Holt
    Sep 2, 2022 at 18:13
  • $\begingroup$ I'm very sorry. $\endgroup$ Sep 2, 2022 at 18:24

1 Answer 1

2
$\begingroup$

There are examples with $A = C_3 \times C_5\ (\cong C_{15})$ and $G = D_8$ (dihedral of order $8$).

We have $A = \langle x,y \mid x^3=y^5=1,xy=yx \rangle$ and $G = \langle a,b \mid a^4=1, b^2=1, b^{-1}ab=a^{-1} \rangle$.

We define $\alpha$ and $\beta$ by $$ \alpha(a)(x)=x, \alpha(a)(y)=y^{-1}, \alpha(b)(x)=x^{-1}, \alpha(b)(y)=y,$$ $$ \beta(a)(x)=x^{-1}, \beta(a)(y)=y, \beta(b)(x)=x, \beta(b)(y)=y^{-1}.$$

Then $H:=\ker(\alpha) = \ker(\beta) = \langle a^2 \rangle$ and $G/H \cong C_2 \times C_2$ is abelian. It is clear that $\alpha(G) = \beta(G)$.

The two semidirect products are not isomorphic - they have different numbers of elements of order $6$ for example. In the GAP and Magma libraries they are $\mathtt{SmallGroup}(120,13)$ and $\mathtt{SmallGroup}(120,12)$.

$\endgroup$
3
  • $\begingroup$ Thank you very much sir. Can you explain why they have different numbers of elements of order 6. $\endgroup$ Sep 3, 2022 at 13:17
  • $\begingroup$ I found by my hand that the first semi direct product has 2 element of order 6 and the second has 22 element of order 6. Is this computation true. $\endgroup$ Sep 4, 2022 at 14:38
  • $\begingroup$ Yes that's right! $\endgroup$
    – Derek Holt
    Sep 4, 2022 at 16:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .