Nearly everyone got this right using Euclid's algorithm.
One solution is 1 = 3×61-7×26 (though there are other solutions).
2.
A common error was to assume that
if the pi's are prime p1×¼×pn+1 is prime.
It is not always; however any prime factor of it must be a new prime.
3.
A common error was to put all rotations in the same conjugacy class.
There are 6 conjugacy classes as follows:
1.
The identity element.
2.
Two rotations by 1/6 of a revolution.
3.
Two rotations by 1/3 of a revolution.
4.
One rotation by 1/2 a revolution.
5.
Three reflections in a diagonal through a corner.
6.
Three reflections in a diagonal through a side.
4.
The elements of Un are represented
by 1,2,4,7,8,11,13,14 of orders 1,4,2,4,4,2,4,2. A common error
was to compute the order of g as the smallest integer > 1
such that gn = g, rather than gn = identity element.
Several people used a rather hard method of finding the orders,
by (say) working out 134 and then dividing this by 15. It is easier to
do this as follows: first note that the order must be 1,2,4,or 8
because it divides the order of G. So we just have to work out
131, 132, 134 mod 15 and so on. This can be done
without hard work as follows:
132
º (-2)2 = 4 mod 15
134
= (132)2 º 42 = 16 º 1 mod 15
so 13 has order 4
(and similarly for the other elements).
The group is not cyclic as it has no element of order 8= order of group.
5.
The cosets are given by multiplying all elements of K by some element of G,
so we get
{1,4} = K,
{2,8} = 2K ,
{7,13} = 7K, and
{11,14} = 11K.
A common error was to add something to K rather than multiplying something
by K.
Any bijection from G/K to Z2×Z2 taking the identity
to the identity is a homomorphism in this particular case
(though of course this is not true for most other groups).
File translated from TEX by TTH, version 2.33. On 2 Oct 1999, 09:29.