Assume that \(K\) is a cyclic group, \(H\) is an arbitrary group, and \(\varphi_1\) and \(\varphi_2\) are homomorphisms from \(K\) into \(\Aut(H)\) such that \(\varphi_1(K)\) and \(\varphi_2(K)\) are conjugate subgroups of \(\Aut(H)\). If \(K\) is infinite, assume in addition that \(\varphi_1\) and \(\varphi_2\) are injective.
Prove by constructing an explicit isomorphism that \(H\rtimes_{\varphi_1}K\cong H\rtimes_{\varphi_2} K\).
Suppose \(\sigma_{\varphi_1}(K)\inverseof{\sigma}=\varphi_2(K)\) so that for some \(a\in\mathbb Z\) we have \(\sigma\varphi_1(k)\inverseof{\sigma}=\varphi_2(k)^a\) for all \(k\in K\). Show that the map \(\psi:H\rtimes_{\varphi_1}K\rightarrow H\rtimes_{\varphi_2}K\) defined by \(\psi((h,k))=(\sigma(h),k^a)\) is a homomorphism. Show \(\psi\) is bijective by construcing a 2-sided inverse.
Solution
Write \(K=\langle k\rangle\). Choose \(\sigma\in\Aut(H)\) such that \[ \sigma\varphi_1(K)\sigma^{-1}=\varphi_2(K). \] Then \(\sigma\varphi_1(k)\sigma^{-1}\) is a generator of the cyclic group \(\varphi_2(K)\), so for some \(a\in\ZZ\), \[ \sigma\varphi_1(k)\sigma^{-1}=\varphi_2(k)^a. \]
1. There is an automorphism \(\alpha:K\to K\), \(\alpha(k)=k^b\), such that\[ \sigma\varphi_1(x)\sigma^{-1}=\varphi_2(\alpha(x)) \qquad(x\in K). \]
If \(K\) is infinite, the injectivity of \(\varphi_1\) and \(\varphi_2\) makes both images infinite cyclic. Since \(\varphi_2(k)^a\) generates \(\varphi_2(K)\), we must have \(a=\pm1\). Take \(b=a\).
Now suppose \(|K|=n<\infty\), and let \(m=|\varphi_2(K)|\). Conjugacy of the two images gives \(|\varphi_1(K)|=m\). Since a cyclic group has a unique subgroup of each order, the kernels \(\ker\varphi_1\) and \(\ker\varphi_2\) are equal, of order \(n/m\). Thus \(\varphi_2(k)^a\) generates \(\varphi_2(K)\), so \(\gcd(a,m)=1\).
We claim that \(a\) may be replaced by an integer \(b\equiv a\pmod m\) with \(\gcd(b,n)=1\). For every prime \(p\mid n\) with \(p\nmid m\), choose a residue \(t_p\pmod p\) for which \(a+mt_p\not\equiv0\pmod p\); such a residue exists because multiplication by \(m\) is invertible modulo \(p\). By the Chinese remainder theorem choose \(t\) realizing all these residues and put \(b=a+mt\). If \(p\mid m\), then \(p\nmid a\) and hence \(p\nmid b\); if \(p\mid n\) but \(p\nmid m\), the choice of \(t\) gives \(p\nmid b\). Therefore \(\gcd(b,n)=1\).
Hence \(k\mapsto k^b\) is an automorphism \(\alpha\) of \(K\). Since \(b\equiv a\pmod m\), we have \(\varphi_2(k)^b=\varphi_2(k)^a\), and therefore the displayed intertwining identity holds on the generator \(k\), hence on all of \(K\).
2. The map\[ \Psi:H\rtimes_{\varphi_1}K\longrightarrow H\rtimes_{\varphi_2}K, \qquad \Psi(h,x)=(\sigma(h),\alpha(x)), \]is an isomorphism.
For \((h,x),(h',x')\in H\rtimes_{\varphi_1}K\), \[ (h,x)(h',x')=(h\varphi_1(x)(h'),xx'). \] Using the intertwining identity from 1, \[ \begin{aligned} \Psi((h,x)(h',x')) &=(\sigma(h\varphi_1(x)(h')),\alpha(xx'))\\ &=(\sigma(h)\,\varphi_2(\alpha(x))(\sigma(h')),\alpha(x)\alpha(x'))\\ &=\Psi(h,x)\Psi(h',x'). \end{aligned} \] Thus \(\Psi\) is a homomorphism. Since both \(\sigma\) and \(\alpha\) are automorphisms, \(\Psi\) is bijective. Explicitly, if \(\alpha^{-1}(k)=k^c\), then \[ \Psi^{-1}(h,x)=(\sigma^{-1}(h),\alpha^{-1}(x)) \] is its two-sided inverse.