transfer kernels and the Schur multiplier - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T13:33:57Z http://mathoverflow.net/feeds/question/53730 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53730/transfer-kernels-and-the-schur-multiplier transfer kernels and the Schur multiplier Franz Lemmermeyer 2011-01-29T18:02:26Z 2011-02-02T20:07:29Z <p>Let $\Gamma$ be a finite $2$-group, and let $G$ be any subgroup of index $2$. Moreover, let Ver$: \Gamma/\Gamma' \to G/G'$ denote the group theoretical transfer, and let $M(\Gamma)$ be the Schur multiplier of $\Gamma$. </p> <p>Is it true that $$| \text{ker}\ {\rm Ver}_{\Gamma/\Gamma' \to G/G'} | \le 2 | M(\Gamma) | \quad ?$$ </p> <p>The answer is positive for a couple of groups with small multiplier, such as cyclic $2$-groups, which have trivial Schur multiplier, the groups of order $8$, or dihedral and quaternion groups. Does current computer technology allow finding a counterexample by going through 2-groups of moderate size?</p> <p><b>Edit.</b> Here's a little bit of background. Let $K$ be a quadratic number field whose 2-class group has type (2,2). It is known that the Hilbert 2-class field $K^1$ has cyclic 2-class group, and that the Galois group $\Gamma$ of the second Hilbert 2-class field is either (2,2) itself, a dihedral, quaternion, or semi-dihedral 2-group. Let $K_j/K$ (j=1, 2, 3) denote the three unramified quadratic extensions inside $K^1$, and let $G_j$ be the Galois groups of $K^2/K_j$. Then $\Gamma$ is quaternion or semi-dihedral if and only if in each of the extensions $K_j/K$, exactly one ideal class of order 2 capitulates (i.e., becomes principal). Now these are exactly the groups among the possible Galois groups with trivial Schur multiplier. On the other hand, the transfer of ideal classes corresponds, via Artin's reciprocity law, to the transfer map (Verlagerung) from the abelianization of $\Gamma$ to that of the $G_j$. Thus in this case, we find that the order of the Schur multiplier is equal to one half of the maximal order of the capitulation kernels. </p> http://mathoverflow.net/questions/53730/transfer-kernels-and-the-schur-multiplier/53843#53843 Answer by Jack Schmidt for transfer kernels and the Schur multiplier Jack Schmidt 2011-01-31T00:53:38Z 2011-01-31T00:53:38Z <p>This is just a fairly literal translation of your question into GAP code.</p> <p>If H is a subgroup of index 2 in G, and t is an element of G not in H, then the transfer map from G to H/[H,H] takes a particularly simple form:</p> <p>If h in H, then h⋅1 = 1⋅h and h⋅t = t⋅h<sup>t</sup>, so tr(h) = h⋅h<sup>t</sup>.</p> <p>If g in G∖H, then g⋅1 = t⋅(t<sup>−1</sup>g) and g⋅t = 1 ⋅(gt), so tr(g) = gt t<sup>−1</sup>g = g<sup>2</sup>.</p> <p>In the GAP language, one creates the homomorphism with:</p> <pre><code>tr := function(g,h) local t, k, gen, img; t := First( g, t -&gt; not t in h ); k := MaximalAbelianQuotient(h); gen := GeneratorsOfGroup(g); img := List( gen, function(x) if x in h then return Image( k, x*x^t ); else return Image( k, x*x ); fi; end ); return GroupHomomorphismByImages( g, Image(k), gen, img ); end; </code></pre> <p>We can test the conjecture on 2-groups with the following GAP function:</p> <pre><code>isCounter := g -&gt; not ForAll( MaximalSubgroupClassReps( g ), h -&gt; Index( Kernel( tr( g, h ) ), DerivedSubgroup( g ) ) &lt;= 2 * Product( AbelianInvariantsMultiplier( g ) ) ); </code></pre> <p>We then look for examples of small orders with the following GAP function:</p> <pre><code>OneSmallGroup( [2,4,8,16,32], isCounter ); # 5 seconds </code></pre> <p>which simply returns fail since no such counterexample of order 2, 4, 8, 16, or 32 exists. To test the higher orders one uses:</p> <pre><code>OneSmallGroup( [ 64], isCounter ); # 30 seconds OneSmallGroup( [128], isCounter ); # 10 minutes </code></pre> <p>For larger orders, one needs to be more careful calculating the Schur multiplier, or applying a little theory since the naive method is dealing with larger index subgroups and then larger integer matrices.</p> http://mathoverflow.net/questions/53730/transfer-kernels-and-the-schur-multiplier/53863#53863 Answer by Chris Gerig for transfer kernels and the Schur multiplier Chris Gerig 2011-01-31T05:56:24Z 2011-02-01T02:28:20Z <p>Take the short exact sequence of modules $\mathbb{Z}\hookrightarrow Ind^\Gamma_G\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ and apply $H_i(\Gamma,-)$. Note that this sequence is exact because $|\Gamma:G|=2$. You obtain the long exact sequence, noting that $H_i(\Gamma,Ind^\Gamma_H\mathbb{Z})\cong H_iG$ by Shapiro's Lemma. As I remarked in the comment attached to this post, the coefficient module from the latter $\mathbb{Z}$ in the short exact sequence has nontrivial $\Gamma$-action (I will denote this coefficient by $\tilde{\mathbb{Z}}$, where the action is multiplication by $-1$ via elements of the nontrivial coset of $\Gamma/G$). Keeping that in mind, we have in particular:</p> <p>$H_2(\Gamma,\tilde{\mathbb{Z}})\stackrel{\delta}{\rightarrow}H_1\Gamma\stackrel{tr}{\rightarrow}H_1G\rightarrow H_1(\Gamma,\tilde{\mathbb{Z}})$.</p> <p>Exactness implies $Ker(tr)=Im(\delta)=H_2(\Gamma,\tilde{\mathbb{Z}})/Ker(\delta)$, so that $|Ker(tr)|\le |H_2(\Gamma,\tilde{\mathbb{Z}})|$. And we know that $tr=Ver$ and $H_2\Gamma=M(\Gamma)$.</p> <p>I want to claim that $|H_2(\Gamma,\tilde{\mathbb{Z}})|\le 2\cdot|H_2\Gamma|$ (the latter homology has $\mathbb{Z}$-coefficient with trivial action), but at this moment I am unsure how to prove it. Hatcher's Algebraic Topology textbook gives a long exact sequence for general coefficient systems (pg330), with $H_3G\stackrel{res}{\rightarrow}H_3\Gamma\rightarrow H_2(\Gamma,\tilde{\mathbb{Z}})\rightarrow H_2G\stackrel{res}{\rightarrow}H_2\Gamma$, so this could be of use. </p> <ul> <li>Chris Gerig</li> </ul>