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Let $K$ be a cyclic group. Let $\phi,\psi: K\rightarrow Aut(H)$ be group homomorphisms such that there exists $\zeta\in Aut(H)$ satisfying $\phi(K)=\zeta \psi(K)\zeta^{-1}$. Then can we prove $H\rtimes_{\phi}K\simeq H\rtimes_{\psi}H$?

My idea is to use the following

Theorem: Let $K,H$ be groups. Let $\phi,\psi: K\rightarrow Aut(H)$ be group homomorphisms and $\zeta\in Aut(H)$, $\alpha\in Aut(K)$ and $\psi=\sigma_{\zeta}\circ\phi\circ\alpha$, where $\sigma_{\zeta}\circ \phi\circ\alpha(k)=\zeta(\phi(\alpha(k)))\zeta^{-1}$ for all $k\in K$. Then $H\rtimes_{\phi}K\simeq H\rtimes_{\psi}H$.

Let $K=\langle a\rangle$ generated by $a$. From the condition $\phi(K)=\zeta \psi(K)\zeta^{-1}$, we have $\phi(a)=\zeta \psi(a^n)\zeta^{-1}$ for some integer $n$. Then how to continue? Thanks.

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4 Answers 4

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The problem is that the action of $\zeta$ might not lift to the whole of $K$.

As an example, let $H=\langle z \mid z^7=1 \rangle = C_7$ and $K = \langle x,y \mid x^3=y^7=1,x^{-1}yx=y^2 \rangle$ be the nonabelian group of order $21$.

Define $\phi,\psi$ by $\phi(x)= (z \mapsto z^2), \phi(y)= 1$, and $\psi(x)= (z \mapsto z^4), \psi(y)= 1$.

So $\phi(K)=\psi(K)$ and we can take $\zeta=1$.

The two semidirect products are

$\langle x,y,z \mid x^3=y^7=z^7=1,yz=zy,x^{-1}yx=y^2,x^{-1}zx=z^2 \rangle$, and

$\langle x,y,z \mid x^3=y^7=z^7=1,yz=zy,x^{-1}yx=y^2,x^{-1}zx=z^4 \rangle$,

which are not isomorphic.

Added later: I realize that this is not a counterexample to the question asked, because $K$ is not cyclic. But I think the example is interesting, so I will not delete it.

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  • $\begingroup$ It's nothing to do with opinions! The problem is that $K$ is not cyclic in my example! I got confused, because I would normally denote the normal subgroup by $K$ and the non-normal subgroup by $H$, and the OP has done the oppopsite. $\endgroup$
    – Derek Holt
    Jan 2, 2014 at 10:40
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If $$\zeta \varphi(a)\zeta^{-1} = \psi(a)^t$$ I believe the map $$ (h, a^i) \mapsto (h^{\zeta}, a^{it} ) $$ provides the required isomorphism from the semidirect product with respect to $\varphi$ to the one with respect to $\psi$. Here I am writing morphisms as exponents.

(Here I am only addressing the first question, with $K = \langle a \rangle $ cyclic.)

Addendum Having seen the second answer by Derek Holt, I now realise this is only a morphism, which should be an isomorphism only when $a \mapsto a^t$ is.

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  • $\begingroup$ I think that every automorphism of a quotient of a finite cyclic group lifts to an automorphism of the whole group, so the result is true for finite cyclic groups $K$. $\endgroup$
    – Derek Holt
    Jan 2, 2014 at 14:26
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Second attempt. I believe this is false when $K$ is infinite cyclic, because automorphisms of finite quotients of $K$ do not usually lift to automorphisms of $K$. For example, the two groups

$\langle x,y \mid x^{11}=1, y^{-1}xy=x^4 \rangle$ and $\langle x,y \mid x^{11}=1, y^{-1}xy=x^5 \rangle$

(with $H = \langle x \rangle$ and $K = \langle y \rangle$) are not isomorphic, but the images of $\phi$ and $\psi$ are both equal to subgroup of ${\rm Aut}(H)$ of order $5$.

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  • $\begingroup$ Thanks, and I still have some questions.(1)What do you mean by "finite quotient of $K$"? (2)Since $ζ\in Aut(H)$, what do you mean by "action of $ζ$ might not lift to the whole of $K$"? $\endgroup$
    – Andrews
    Feb 15, 2019 at 16:26
  • $\begingroup$ A finite quotient of $K$ is a finite quotient group $Q := K/N$ for some normal subgroup $N$ of $K$. For an automorphism $\zeta$ of $Q$, it is possible that there is no automorphism $\theta$ of $K$ with $\theta(N) = N$ such that $\theta(g)N = \zeta(gN)$ for all $g \in K$. In that case we say that $\zeta$ does not lift to an automorphism of $K$. $\endgroup$
    – Derek Holt
    Feb 15, 2019 at 17:43
  • $\begingroup$ So in your answer,$H=<x>\cong \Bbb Z_{11},K=<y>\cong \Bbb Z$, you mean treat $H$ as finite quotient of $K$ and $\phi_1(y)(x)=x^4, \phi_2(y)(x)=x^5$, $\phi_1,\phi_2$ are automorphism of $H$ but can't lift to $K$, is that right? And why that's related to the initial proposition $H⋊_ϕK≃H⋊_ψH$ may not correct? $\endgroup$
    – Andrews
    Feb 16, 2019 at 2:22
  • $\begingroup$ No, the finite quotient of $K$ in question is a subgroup of ${\rm Aut}(H)$ of order $5$. This subgroup is the image of both $\phi$ and $\psi$. I left the proof of non-isomorphism of the two semidirect products as an exercise. $\endgroup$
    – Derek Holt
    Feb 16, 2019 at 10:06
  • $\begingroup$ Thanks so much for your help these days. $H=\langle x\rangle\cong \Bbb Z_{11}, K=\langle y \rangle \cong \Bbb Z. \text{Aut}(H)=\Bbb Z_{11}^\times \cong \Bbb Z_{10}.$ $\phi_1(y)(x)=x^4, \phi_2(y)(x)=x^5, \phi_1(K)=\phi_2(K)=\{\bar 1, \bar3, \bar4, \bar5, \bar9\}\cong \Bbb Z_5$. ker$\phi_1 \cong$ ker$\phi_2=5\Bbb Z$. After some calculation I can derive that $H\rtimes_{\phi_1} K \not \cong H\rtimes_{\phi_2} K$, but I still don't know what is "the" automorphism of $\phi_1(K)=\phi_2(K)\cong \Bbb Z_5$ which can't lift to the whole $K \cong \Bbb Z$. Could you please point it out? Thanks. $\endgroup$
    – Andrews
    Feb 16, 2019 at 15:21
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I think, this is false. Take $H=Z^5$, $K$ cyclic of order 5 generated by $k$, $\phi(k)$ an element $c$ of order 5 and $\psi(k)=c^2$. I think, the corresponding semidirect products are not isomorphic since $c$ is not conjugate to $c^2$ in the group $GL(5,Z)$.

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  • $\begingroup$ I don't think this works. The two semidirect products are isomorphic with the map inducing identity on $H$ and mapping $c$ to $c^2$. I think the result is true whenever $\phi$ and $\psi$ are both injective. $\endgroup$
    – Derek Holt
    Jan 2, 2014 at 9:25
  • $\begingroup$ @Derek Holt: What you defined is not a homomorphism. $\endgroup$ Jan 2, 2014 at 16:14
  • $\begingroup$ Sorry, it should be $c \mapsto c^3$ for an isomorphism $H\rtimes_\phi K \to H \rtimes_\psi K$. Using $\circ_\phi$ and $\circ_\psi$ for composition in the two semidirect products, we get $(h_1,k^i) \circ_\phi (h_2,k^j) = (h_1c^i(h_2),k^{i+j}) \mapsto (h_1c^i(h_2),k^{3(i+j)}) = (h_1,k^{3i})\circ_\psi (h_2,k^{3j})$. $\endgroup$
    – Derek Holt
    Jan 2, 2014 at 17:23

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