Conjugate actions and isomorphic semidirect products

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.

• 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. Sep 2, 2022 at 17:07
• I'm very sorry. Sep 2, 2022 at 18:24

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 $$\mathtt{SmallGroup}(120,12)$$.

• Thank you very much sir. Can you explain why they have different numbers of elements of order 6. Sep 3, 2022 at 13:17
• 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. Sep 4, 2022 at 14:38
• Yes that's right! Sep 4, 2022 at 16:07