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<tdstyle="vertical-align: top;"><a
href="aboutDavisComplex.html"><smallstyle="color: rgb(0, 0, 102);">Previous</small></a><br>
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<td style="text-align: center; vertical-align: top; color: rgb(0, 0, 102);"><big><span style="font-weight: bold;">About HAP: Functorial Properties<br>
</span></big></td>
<tdstyle="text-align: right; vertical-align: top;"><a
href="aboutCoefficientSequence.html"><small style="color: rgb(0, 0, 102);">next</small></a><br>
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<big><spanstyle="font-weight: bold;"></span></big><br>
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<td style="vertical-align: top; background-color: rgb(255, 255, 255); text-align: left;">Homology
is
a
functor.
That is, for any n>0 and group homomorphism <span style="font-family: opensymbol;"></span>f : G → G' there is an style="font-style: italic;">induced
homomorphism</span> H<sub>n</sub>(f) : H<sub>n</sub>(G,Z) → H<sub>n</sub>(G',Z)
satisfying<br>
<ul>
<li>H<sub>n</sub>(gf) = H<sub>n</sub>(g)H<sub>n</sub>(f) for
group homomorphisms f : G → G', g : G' → G".
<li>H<sub>n</sub>(f) is the identity homomorphism if f is the
identity.</li>
</ul>
<spanstyle="font-style: italic;"></span>The following
commands compute H<sub>3</sub>(f) : H<sub>3</sub>(P,Z) → H<sub>3</sub>(S<sub>5</sub>,Z)
for
the
inclusion
f : P → S<sub>5 </sub>into the symmetric group S<sub>5</sub>
of its Sylow 2-subgroup. They also show that the image of the induced
homomorphism H<sub>3</sub>(f) <spanstyle="font-style: italic;"></span>is
precisely
the
Sylow
2-subgroup of H<sub>3</sub>(S<sub>5</sub>,Z).<br>
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<td style="background-color: rgb(255, 255, 204); vertical-align: top;">gap>
S_5:=SymmetricGroup(5);;
P:=SylowSubgroup(S_5,2);;<br>
<br>
gap> f:=GroupHomomorphismByFunction(P,S_5, x->x);;<br>
<br>
gap> R:=ResolutionFiniteGroup(P,4);;<br>
<br>
gap> S:=ResolutionFiniteGroup(S_5,4);;<br>
<br>
gap> ZP_map:=EquivariantChainMap(R,S,f);;<br>
<br>
gap> map:=TensorWithIntegers(ZP_map);;<br>
<br>
gap> Hf:=Homology(map,3);;<br>
<br>
gap> AbelianInvariants(Image(Hf));<br>
[2,4]<br>
<br>
gap> GroupHomology(G,3);<br>
[2,12]<br>
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<td style="background-color: rgb(255, 255, 255); vertical-align: top;">The
above computation illustrates a general result.<br>
<br>
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<td style="text-align: left; background-color: rgb(204, 255, 255); vertical-align: top;"><span style="color: rgb(0, 0, 102);">For any Sylow p-subgroup P of a finite
group G, the p-primary part of H</span><sub style="color: rgb(0, 0, 102);">n</sub><span style="color: rgb(0, 0, 102);">(G,Z) is a quotient of H</span><sub style="color: rgb(0, 0, 102);">n</sub><span style="color: rgb(0, 0, 102);">(P,Z).</span> </td>
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<br>
We denote by <spanstyle="color: rgb(0, 0, 102);">H</span><sub style="color: rgb(0, 0, 102);">n</sub><span style="color: rgb(0, 0, 102);">(G,Z)<sub>(p)</sub> the p-part of </span><span style="color: rgb(0, 0, 102);">H</span><sub style="color: rgb(0, 0, 102);">n</sub><span style="color: rgb(0, 0, 102);">(G,Z). </span>This result follows from
a property of the <spanstyle="font-style: italic;">transfer
homomorphism <br>
<br>
</span>
<divstyle="text-align: center;">Tr(G,K) : H<sub>n</sub>(G,Z) → H<sub>n</sub>(K,Z)
<br>
</div>
<br>
which exists for any group G and subgroup K<G of finite index |G:K|,
and any n>0. The relevant property is the following.<br>
<br>
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<td style="background-color: rgb(204, 255, 255); vertical-align: top;">The
composed homomorphism <br>
<br>
<divstyle="text-align: center;">H<sub>n</sub>(K<G)oTr(G,K)
:
H<sub>n</sub>(G,Z) → H<sub>n</sub>(G,Z)<br>
</div>
<br>
is just multiplication by the index |G:K|.<br>
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<br>
So in the case when K is a Sylow p-subgroup the composed homomorphism
is an
isomorphism on H<sub>n</sub>(G,Z)<sub>(p)</sub> and, consequently, the
induced homomorphism H<sub>n</sub>(K→G,Z) must map surjectively
onto H<sub>n</sub>(G,Z)<sub>(p)</sub>.<br>
<br>
Another consequence of the transfer (with K=1) is that the exponent of H<sub>n</sub>(G,Z)
divides
the
order
|G| for any finite group G. In particular, H<sub>n</sub>(G,Z)
is finite (since it is readily seen to be a finitely
generated abelian group).<br>
<br>
These results suggest that the homology H<sub>n</sub>(G,Z) of a large
finite group G (such as the Mathieu group G=M<sub>23</sub>) should be
calculated by computing its p-part H<sub>n</sub>(G,Z)<sub>(p)</sub>
for each prime p dividing |G|. For a Sylow p-subgroup P there is a nice
description of the kernel of the surjection H<sub>n</sub>(P,Z) → H<sub>n</sub>(G,Z)<sub>(p)</sub>.
It
is
generated
by
elements <br>
<br>
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<td style="background-color: rgb(204, 255, 255); text-align: center; vertical-align: top;"><span style="color: rgb(0, 0, 102);">h</span><sub style="color: rgb(0, 0, 102);">K</sub><span style="color: rgb(0, 0, 102);">(a)
- h</span><substyle="color: rgb(0, 0, 102);">xKx<sup>-1</sup></sub><span style="color: rgb(0, 0, 102);">(a)</span><br>
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<br>
where x ranges over the double coset representatives of P in G, K is
the intersection of P and its conjugate xPx<sup>-1</sup>, the
homomorphisms h<sub>K</sub>, h<sub>x<sup>-1</sup>Kx</sub>:H<sub>n</sub>(K,Z)
→
H<sub>n</sub>(P,Z) are induced by the inclusion K→P,
k→k and the conjugated inclusion K→P, k→x<sup>-1</sup>kx,
and a ranges over the generators of H<sub>n</sub>(K,Z).<br>
<br>
The function <spanstyle="font-family: helvetica,arial,sans-serif;">PrimePartDerivedFunctor(G,R,T,n)</span>
uses this
description of the kernel to compute the abelian invariants
of H<sub>n</sub>(G,Z)<sub>(p)</sub> starting from: <br>
<ul>
<li>the group G, </li>
<li>a ZP-resolution R for a Sylow p-subgroup P<G, </li>
<li>the functor T=TensorWithIntegers,</li>
<li>the integer n>0.<br>
</li>
</ul>
The following commands show that the Mathieu group M<sub>23</sub> has
third integral homology with 2-part H<sub>3</sub>(M<sub>23</sub>,Z)<sub>(2)</sub>=0.
<br>
</td>
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<td style="background-color: rgb(255, 255, 204); vertical-align: top;">gap>
M_23:=MathieuGroup(23);;<br>
<br>
gap> gens:=GeneratorsOfGroup(SylowSubgroup(M_23,2));;<br>
<br>
gap> gensP:=[gens[4],gens[6],gens[2]];; #This list
generates a Sylow 2-subgroup of M_23.<br>
<br>
gap> R:=ResolutionFiniteGroup(gensP,4);;<br>
<br>
gap> T:=TensorWithIntegers;;<br>
<br>
gap> PrimePartDerivedFunctor(M_23,R,T,3);<br>
[ ]<br>
</td>
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<td style="text-align: left; background-color: rgb(255, 255, 255); vertical-align: top;"><a
name="Milgram"></a>Similar
commands can be used to show that H<sub>n</sub>(M<sub>23</sub>,Z)=0 for
n=1,2,3,4. The triviality of the first four integral homology
groups of M<sub>23</sub> was first proved in [J. Milgram, J. Group
Theory, 2000] and answered a conjecture of J.-L. Loday. A group G is
said to be <spanstyle="font-style: italic;">k-connected</span> if H<sub>n</sub>(G,Z)=0
for
n=1,
2,
..., k. Back in the mid 1970s Loday had asked if the
trivial group is the only 3-connected finite group. <br>
<br>
No example of a 5-connected finite group is yet known! A group is said
to be <spanstyle="font-style: italic;">superperfect</span> if it is
2-connected. A list of some superperfect groups, together with their
third integral homology, is given <a style="font-weight: bold;"
href="aboutSuperperfect.html">here</a>.<br>
<br>
The higher dimensional integral homology of a group G is readily
calculated by this method when G has no large Sylow subgroup. For
example, the following commands show that the symmetric group of degree
5 has 20-dimensional integral homology H<sub>20</sub>(S<sub>5</sub>,Z)
= (Z<sub>2</sub>)<sup>7</sup> . (We could of course have incorporated
into our computation the fact
that cyclic Sylow groups have trivial integral homology in even
dimensions.)<br>
</td>
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<td style="background-color: rgb(255, 255, 204); vertical-align: top;">gap>
S_5:=SymmetricGroup(5);;
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