Assignment:

1.  Let G be a group of order 56. 

(a)  How many 2-sylow and 7-sylow subgroups are there.  

         1 2 4 7 8 14 28 56
2-sylow  x     x
7-sylow  x       x


(b)  If there are 8 7-sylow subgroups, how many elements of 

order 7 are there.  

In any 7-sylow subgroup with 7 elements, all the elements

except the identity are of order 7.  Therefore there are

6 elements of order 7 in each 7-sylow subgroup.  Since 7

is a prime, none of these elements of order 7 can be in

more than one 7-sylow subgroup.  Thus the number of elements

of order 7 is 8x7 = 48. 

(c)  If there are 6 7-sylow subgroups, how many elements are

there which are not of order 7.  Is there enough to make more

than one 2-sylow subgroup?

There are 56-48 = 8 elements which are not of order 7.  This

is only enough for one 2-sylow subsgroup.


2.  Let H be a subgroup of the group G.  Show that

N(H) = {g in G | gHg' = H} is a subgroup of G.

We have to show closure and inverses.

closure:  Let g1 and g2 elements in N(H).

(g1g2)H(g1g2)' = (g1g2)H(g2'g1') = g1(g2Hg2')g1' = g1Hg1' = H.

inverse:  Let g be an element of N(H).

g'H(g')' = g'Hg = g'(gHg')g = (g'g)H(g'g) = H.

   
                                                  r
3. Show that there are no simple groups of order p m where

p is prime and m < p.

I will show that the p-sylow subgroup is normal.  

The number of p sylow subgroups divides the order of the

group so the number of them is of the form p^s m.  

0 <= s <= r and m divides n.  The number of p sylow subgroups 

is also equal to one mod p.  Therefore s = 0.  Since

1 <= n <= m < p, the only possible number for n is 1.

Therefore there is only one p-sylow subgroup and it

is normal.