Problems:  
1(a). How many 6 bead necklaces are there using exactly 2 red, 2 blue,
      and 2 green beads?

     1    I                  90
     2 (******)               0
     2 (***)(***)             0
     1 (**)(**)(**)           6

    3  (*)(*)(**)(**)        18
    3  (**)(**)(**)          18
                           -------
                            132/12 = 11 <== ans



 (b). How many 8 bead necklaces are there using exactly 3 red and 5
      blue beads?

1  I                           1 56 =  56
4 (********)                   4  0
2 (****)(****)                 2  0
1 (**)(**)(**)(**)             1  0
  
4 (**)(**)(**)(**)             4  0
4 (*)(*)(**)(**)(**)           4  6  = 24
--                                  ------
16                                     80/16 =  5  <====ans

 (c).  How many 8 bead necklaces are there using exactly 4 red and
       4 blue beads?


1  I                           1  70 = 70 
4 (********)                   4   0     
2 (****)(****)                 2   2 =  4 
1 (**)(**)(**)(**)             1   6 =  6 
  
4 (**)(**)(**)(**)             4   6 = 24
4 (*)(*)(**)(**)(**)           4   6 = 24 
--                                  ------
16                                    128/16 = 8 




2.  Suppose that G is a finite group.  Let H be a subgroup of G.

                                            -1 
    For fixed element g of G, prove that gHg   is also a subgroup

    of G.


Proof: We have to show closure and inverses.

If g h1 g' and g h2 g' are in gHg', then

(g h1 g')(g h2 g') = g h1 h2 g'  which is in gHg' since h1 h2 is in H.

(g h g')' = g'' h' g' = g h' g' since g'' = g.  Thus

the inverse of g h g' is also in gHg' so gHg' is a subgroup.



3.  Let S be the collection of all subgroups of G.  Show that
 
the action * of G acting of S where if g is in G and H is in S,

            -1
then g*H = g  H g  satisfies the requirements of a group acting

                           -1    -1 -1
on a set.  Hint:  Show (ab)   = b  a  .


Proof:  There are two things we have to show. 

(1)  e*s = s

(2)  (g1 g2)*s = g1*(g2*s)).


e*H = e' H e = e H e = H.

(g1 g2)*H = (g1 g2)' H (g1 g2) = g2' g1' H g1 g2 = g2'(g1' H g1) g2

          = g2*(g1*h). 

Actually, this does not work correctly.  I should have said

           -1
g*H = g H g

Now

(g1 g2)*H = (g1 g2) H (g1 g2)' = (g1 g2) H (g2' g1') =

g1(g2 H g2')g1' = g1*(g2*H).






4.  How many elements are there in the group of symmetries of:
(a) the equilateral triangle,    ans 6
(b) the square,                  ans 8
(c) the pentagon,                ans 10
(d) the octahedron,              ans 24
(e) the dodecahedron.            ans 60