Conjugation of group extensions

Question

Let $H$ be a finite group. We write ${{\mathbb{C}}}^{*n}$ for the $n$-dimensional complex torus $({{\mathbb{C}}}^*)^n$. We have a short exact sequence $$ 0\to {{\mathbb{Z}}}^n\to {{\mathbb{C}}}^n\to{{\mathbb{C}}}^{*n}\to 1,$$ which gives a connecting isomorphism $$ \Delta\colon {\rm Hom}(H,{{\mathbb{C}}}^{*n})\to H^2(H,{{\mathbb{Z}}}^n). $$

Let $\varphi\colon H\to {{\mathbb{C}}}^{*n}$ be a homomorphism. We obtain a central group extension $$ 0\to {{\mathbb{Z}}}^n\to E(\varphi)\to H\to 1$$ corresponding to $\Delta(\varphi)\in H^2(H,{{\mathbb{Z}}}^n)$. Namely, $$ E(\varphi)=H\times_{{{\mathbb{C}}}^{*n},\varphi} {{\mathbb{C}}}^n.$$

Now let $\sigma\in {\rm Aut}({{\mathbb{C}}})$ be an automorphism (not necessarily continuous). We define an action of $\sigma$ on ${{\mathbb{C}}}^{*n}$ by $\sigma(z_1,\dots,z_n)=(\sigma(z_1),\dots,\sigma(z_n))$. We obtain a new homomorphism $$ \sigma\varphi:=\sigma\circ\varphi\colon\ H\to {{\mathbb{C}}}^{*n}$$ and a new extension $$ 0\to {{\mathbb{Z}}}^n\to E(\sigma\varphi)\to H\to 1,$$ which is not isomorphic to $E(\varphi)$ when $\sigma\varphi\neq\varphi$.

Question. Can it happen that $E(\sigma\varphi)$ is not isomorphic to $E(\varphi)$ as groups (and not only as group extensions)?

Motivation: I am trying to construct a homogeneous space $G/H$ of a connected linear ${{\mathbb{C}}}$-group $G$, such that the topological fundamental groups of $\tau(G/H)$ and of $G/H$ are not isomorphic.

Answer 1

Here's an example where they are not isomorphic, with $n=1$.

Let $k$ be the order of the image of $\varphi$, which is also the image of $\sigma\phi$ (since $\mathbf{C}^*$ has a single subgroup of each given order). Then $\varphi$ can be viewed, after some fixed identification, as the fibre product $H\times_{(\mathbf{Z}/k\mathbf{Z})^n,\varphi}\mathbf{Z}^n$. An element $\sigma$ is just the choice of some element in $\ell\in (\mathbf{Z}/k\mathbf{Z})^*$, and this changes $\varphi$ to $\varphi^\ell:h\mapsto\varphi(h)^\ell$, .

Now consider the semidirect product $H=(\mathbf{Z}/5\mathbf{Z})\ltimes_4 (\mathbf{Z}/11\mathbf{Z})$, here the notation means that the generator of order 5 acts by multiplication by 4 on the right (which is indeed of order 5 modulo 11), and $\varphi$ the canonical projection to $\mathbf{Z}/5\mathbf{Z}$, and $\sigma_c$ the automorphism of $\mathbf{Z}/5\mathbf{Z}$ given by multiplication by $c\in(\mathbf{Z}/5\mathbf{Z})^*$. Then $E(\sigma_c\varphi)=\mathbf{Z}\ltimes_{4^{c^{-1}}}(\mathbf{Z}/11\mathbf{Z})$, where the generator of $\mathbf{Z}$ acts by multiplication by $4^{c^{-1}}$ (beware that $c^{-1}$ is modulo 5 but $4^{c^{-1}}$ is modulo 11). Then $E(\sigma_1\varphi)=E(\varphi)$ and $E(\sigma_2\varphi)$ are not isomorphic ($4^{2^{-1}}$ is then equal to $-2$). The argument if that these groups both have a unique normal subgroup of order 11, and a unique homomorphism $f$ onto $\mathbf{Z}$ up to sign, and that elements of $f^{-1}(\{-1,1\})$ act on the normal subgroup by multiplication by $\pm 4$ in $E(\varphi)$, by multiplication by $\pm 2$ in $E(\sigma_2\varphi)$, and the automorphisms of $\mathbf{Z}/11\mathbf{Z}$ of multiplication by $\pm 2$ and by $\pm 4$ are not conjugate.

In your original language, $\varphi$ is the homomorphism to $\mathbf{C}^*$ mapping the generator of order 5 to $\omega_5=\exp(2i\pi/5)$ and mapping the generator of order 11 to 1, and $\sigma$ is any field automorphism of $\mathbf{C}$ mapping $\omega_5$ to $\omega_5^2$.