From: lounesto@dopey.hut.fi (Pertti Lounesto) Newsgroups: sci.math Subject: Q: Extension of automorphism of a subgroup Date: 20 Jan 1995 15:07:04 GMT Let G be a group, and H its subgroup, and a:H->H an automorphism of H. Is there always an automorphism of G, whose restriction to H is a, and if not, could you give a counter-example for the finite group (of lowest order)? -- Pertti Lounesto E-Mail: lounesto@dopey.hut.fi Helsinki Univ. of Technology Newsfeed unreliable. I might have missed the beginning of this thread, so forgive me if I'm repeating facts in posts that have expired on my system. If you post a follow-up to this article, send it to me by E-mail, as well. ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Q: Extension of automorphism of a subgroup Date: 20 Jan 1995 16:47:35 GMT In article , Pertti Lounesto wrote: >Let G be a group, and H its subgroup, and a:H->H an automorphism of H. >Is there always an automorphism of G, whose restriction to H is a, and >if not, could you give a counter-example for the finite group (of lowest >order)? Let G be the dihedral group of order 8 (the group of symmetries of the square). It has an elementary-abelian subgroup H of order 4 generated by the reflections across the edges of the square. Now, H itself has 6 automorphisms (arbitrarily permute the 3 elements of order 2). But not all of these extend to the whole group G: H includes the 180-degree rotation of the square, which is in the center of G; the other involutions of H are not in the center of G, so no automorphism of G can permute these. dave ============================================================================== From: lounesto@dopey.hut.fi (Pertti Lounesto) Newsgroups: sci.math Subject: Re: Q: Extension of automorphism of a subgroup Date: 21 Jan 1995 14:24:57 GMT On 20 Jan 1995 rusin@washington.math.niu.edu (Dave Rusin) gave a nice answer to my question, thank you. Your response in fact resolved a problem I had for a while, and your counter-example works as such also in my much more abstract case. Here was my problem: Does the triality automorphism of Spin(8) extend to Pin(8)? Triality permutes the three non-trivial elements {-1, e12...8, -e12...8} in the center of Spin(8), but only -1 is in the center of Pin(8). So your counter- example shows that triality does not extend to Pin(8). Dave Rusin, thank you once more for your nice resolution. -- Pertti Lounesto E-Mail: lounesto@dopey.hut.fi Helsinki Univ. of Technology Newsfeed unreliable. I might have missed the beginning of this thread, so forgive me if I'm repeating facts in posts that have expired on my system. If you post a follow-up to this article, send it to me by E-mail, as well. ============================================================================== Newsgroups: sci.math From: william@math.Princeton.EDU (William Schneeberger) Subject: Re: Q: Extension of automorphism of a subgroup Date: Sat, 21 Jan 1995 21:51:21 GMT In article lounesto@dopey.hut.fi (Pertti Lounesto) writes: >Let G be a group, and H its subgroup, and a:H->H an automorphism of H. >Is there always an automorphism of G, whose restriction to H is a, and >if not, could you give a counter-example for the finite group (of lowest >order)? >-- > Pertti Lounesto E-Mail: lounesto@dopey.hut.fi > Helsinki Univ. of Technology > >Newsfeed unreliable. I might have missed the beginning of this thread, so >forgive me if I'm repeating facts in posts that have expired on my system. >If you post a follow-up to this article, send it to me by E-mail, as well. Someone else pointed out that D4 would be the G of minimal order, with H of order 4. However, we can get H to be of order 3. In the nonabelian group of order 21, the semidirect product of C7 with C3, we can find an element a of order 3 such that for any x of order 7, x^a=x^2, but x^(a^2)=x^4. So the group admits an automorphism which does not extend to the larger group. -- Will Schneeberger Hi There !! william@math.Princeton.EDU ============================================================================== From: LAURAHELEN@news.delphi.com (LAURAHELEN@DELPHI.COM) Newsgroups: sci.math Subject: Re: Q: Extension of automorphism of a subgroup Date: 21 Jan 1995 19:04:09 -0500 >In general a counterexample is a direct sum of abelian groups with the same >order. Any permutation of the generators induces an automorphism of the >direct sum. Now embed this direct sum into a direct sum where one factor >is a cyclic group of higher order. Any permutation of the original >generators which involves the factor that was extended , can't be lifted >to the larger direct sum. I meant "direct sum of cyclic groups with the same order". Anyway, this style of counterexample does not seem to work in the case where G is finite order, H < G, and |H| and |G|/|H| are relatively prime. So I wonder whether an automorphism of H can always be lifted to G in this case. *she goes off to mull* Laura ============================================================================== From: ellenber@husc7.harvard.edu (Jordan Ellenberg) Newsgroups: sci.math Subject: Re: Q: Extension of automorphism of a subgroup Date: 21 Jan 95 22:06:12 LAURAHELEN asked whether, if G is a finite group, H a subgroup of G with |H| and [G:H] relatively prime, an automorphism of H can be lifted to an automorphism of G. I think the answer is still no. Let H be the direct sum of four groups of order 2 and let h1,h2,h3,h4 be the generators of H. Let G be the group of order 48 generated by H and g, where g^3=1 and conjugation by g permutes h2,h3,h4 cyclically. Then G and H are of the desired form, but the automorphism of H transposing h1 and h2 does not lift to G, since h1 is central in G and h2 is not. Jordan Ellenberg ============================================================================== From: LAURAHELEN@news.delphi.com (LAURAHELEN@DELPHI.COM) Newsgroups: sci.math Subject: Re: Q: Extension of automorphism of a subgroup Date: 21 Jan 1995 22:40:16 -0500 LAURAHELEN@news.delphi.com (LAURAHELEN@DELPHI.COM) writes: >Anyway, this style of counterexample does not seem to work in the case >where G is finite order, H < G, and |H| and |G|/|H| are relatively prime. >So I wonder whether an automorphism of H can always be lifted to G in >this case. It can't always. Let H = C_3 x C_3, generated by b and c. Let a be an element of order 2. Let G be the semidirect product of with H; let a act on H by f(a)(b) = b^2, f(a)(c) = c. Then there is an automorphism s of H which exchanges b and c, but it can't be lifted to G because c is in the center of G, while b is not. But if G is abelian then G is the direct product G/H x H and the automorphism can be lifted. Laura