Homework #6, due March 11th:
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Let G = Z_2 x Z_2 x Z_2,
with elements 000,001,010,100,101,011,110,111 added mod 2.
Find three automorphisms of G, of order 2,3, and 7. 
(This is related to the Fano hypergraph I drew at the end of class;
can you see how?) In fact |Aut(G)|=168 and its only prime factors are
2,3,7.
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Order 2 is easy: (a,b,c) -> (b,a,c) works. Order 3 is easy too:
(a,b,c) -> (b,c,a). Order 7 is much trickier.
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The main things to guide the search are that any automorphism 
permutes not only the 7 vertices but the 7 edges. To be of order 7
it's got to therefore either give a 7-cycle of each, or be a 7-cycle
on one while acting trivially on the other, for example being a 7-cycle
on the vertices while taking each line to itself (mixed up). 
But if we take some 3-pt line to itself, we'd see 2- or 3-cycles of the
vertices on that line. So in fact it has to be a 7-cycle of the vertices
and also a 7-cycle of the edges.
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One that works is (001 100 110 111 011 101 010). Try labeling the
vertices of the Fano hypergraph with nonidentity elements of G,
then recopy it but with this permutation applied, to see that lines
still each add up to 000.
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4.1 #9a, part 1.
If O_i = H.a = the H-orbit through a, then g H.a = H.(g.a) 
= the H-orbit through g.a. Ta-da!
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(This is a little unfair; the usual definition of normality says that
for each h in H, g h g^{-1} is in H again. That says that g H g^{-1} 
is part of H, for all g. So apply it for g^{-1}: now g^{-1} H g is part
of H. So H is part of g H g^{-1}. Hence g H g^{-1} = H, and g H = H g.)
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Part 2. To get from H.a to H.b, let g be such that g.a = b. Then
g.(H.a) = gH.a = Hg.a = H.b. So transitivity on A gets transitivity on
this set, H\A.
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Part 3. With notation from part 2, we 
have maps g. : H.a -> H.b and g^{-1}. : H.b -> H.a, and these maps
are inverses. Hence H.a and H.b have the same size.
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4.1 #10a.
This is just fighting with notation. Each left coset (h x) K has elements
{h x k : k in K}. Take the union over H (or equivalently over the
H-orbit), we get {h x k : k in K, h in H}. But that's H x K.
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b. HxK = union_K (H x k)  = union_K H(xk)
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c. Ugh. There are serious notational problems here because I want to 
refer to the group H times K, but H x K is taken already. So write H*K
to mean the Cartesian product group.
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All right, let H*K act on G by (h,k).g = h g k^{-1}. Then the double
cosets are exactly the orbits of H*K on G. Being orbits, any two are either
disjoint or identical. You may have repeated the usual proof of this,
which is fine.
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d,e. Let (H*K)_x be the stabilizer of x in G, i.e. 
{(h,k) : h x k = x}. Rewriting, h = x k x^{-1}. 
So we get a correspondence of (H*K)_x with H intersect x K
x^{-1}. Using our theorem computing the size of orbits, we get 
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|HxK| = |H*K : (H*K)_x|.
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Note that HxK contains H and K, so it's only reasonable to compute its
size if H and K are each finite. Then the above says <br>
|HxK| = |H*K| / |(H*K)_x| = |H| |K| / |H intersect x K x^-1| <br>
or equivalently, <br>
|HxK| = |H| |K| / |x^-1 H x intersect K| <br>
and these give c,d.

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Let G = the rotations of the cube, so |G|=24. Let V = the 8 vertices
of the cube. Let VxV = the set of ordered pairs of elements.
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<li>How many orbits are there of G on VxV? For each orbit, say how many
elements of VxV are in it, write down one of them, and describe the
stabilizer.
<li>Answer the same questions for P = the set of 2-element subsets of V
(note that these are unordered: |P| = 28).
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Ans: VxV. There are four orbits, depending on the distance of (v,w),
which is 0, 1, sqrt(2), or sqrt(3). Their sizes are 8, 24, 24, 8.
In the first and last cases, the stabilizers fix a long diagonal,
leaving three rotations. In the other two cases the stabilizers are trivial.
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Ans: P. There are three orbits, again by distance, but 0 is now impossible.
Their sizes are 12, 12, 4 -- the sizes in VxV modulo switching the two 
elements. In the distance 1 case, the two vertices {v,w} share an edge,
and the stabilizer of {v,w} is {Id, the 180 degree rotation fixing
the midpoint of the edge vw}. In the distance sqrt(2) case, they share a face,
and the stabilizer of {v,w} is {Id, the 180 degree rotation fixing
the center of their common face}. In the distance sqrt(3) case, 
they share a long diagonal, and the stabilizer either rotates that diagonal
or switches the two ends.
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Find a normal subgroup K of S_4, and a normal subgroup H of K, such
that H is not a normal subgroup of S_4. ("Normality isn't transitive.")
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Ans. In fact the only normal subgroups of S_4 are S_4,A_4,{Id}, and one
other. You have to let K be this one other, {Id} union the 2,2-cycles.
(This was the kernel of Rot(cube) acting on pairs of opposite faces.)
Then K is abelian, so any Z_2 inside it is normal in K, but isn't
a union of conjugacy classes up in S_4.