Subgroups

1. Subgroups A subgroup of a group G is a subset of G’s elements which themselves will form a group under the same operation. • {e} and G itself will always be subgroups and these are called trivial subgroups. • All other subgroups are called proper subgroups. Ex 1 Find subgroups of (Z6 , +)

2. Note: All subgroups must contain the identity element. You can usually write down subgroups by trial and error. For example {0, 1, 2} will not be a subgroup because 22 = 4 so {0, 1, 2} won`t be closed. The proper subgroups are {0, 2, 4} and {0, 3} {0} and G itself are trivial subgroups so in all there are 4 subgroups of G.

3. You need to check three things Closure – H is closed under • Identity – The identity of G lies in H Inverse – The inverse of elements of H also lie in H As the original group is associative then the subgroup is also Ex 4 Show that H = (2Z, +) is a subgroup of (Z, +) Z is the set of integers {0, 1,  2,  3….} Two elements of 2Z are 2n and 2m. Closure – 2n + 2m = 2(n + m) but 2(n + m) is an element of H Identity – 0 is the identity of (Z, +) and this also lies in the set H Inverse – The inverse of +2n is –2n which is also even

4. A method for finding all subgroups of a cyclic group is to list elements generated by powers of each element in the table. Only subgroups are {0}, G, {0,4,6} and {0,2}. 01=02=03=…. So 0 generates the trivial subgroup {0}. 21=2, 22=0, 23=2…. So 2 generates the subgroup {0,2}. 31=3, 32=4, 33=2, 34=6, 35=5, 36=0 So 3 generates the trivial subgroup G. 41=4, 42=6, 43=0 So 4 generates the subgroup {0,4,6}. 51=5, 52=6, 53=2, 54=4, 55=3, 56=0 So 5 generates the trivial subgroup G. 61=6, 62=4, 63=0 So 6 generates the subgroup {0,4,6} again.

5. Ex 1) Find all subgroups of {0, 1, 2, 3, 4, 5, 6, 7, 8} under + modulo 9. 2) Find all subgroups of {Z4 , +} 3) Find all subgroups of {Z6 , +} What do you notice about the order of the subgroups and the order of the group? Qu.s 1) The set M {1, 2, 3, 4, 5, 6, 7, 8, 9, 10 } is a group under the binary operation of multiplication modulo 11. List all the proper subgroups of M. 2) The group G = {1, 2, 3, 4, 5, 6} has multiplication modulo 7 as its operation. List all the proper subgroups of G. Ex 10B pg. 399/400 No. 1(pinch table from back of book) 3, 8

6. Lagrange’s Theorem The order of a subgroup is a factor of the order of the group. (The converse doesn’t apply) • Lagrange’s Theorem is helpful when trying to list subgroups. It also explains why groups of prime order can have no proper subgroups. So if the order of the group is 6 then the order of the subgroups are 1,2,3 and 6 The subgroups of order 1 and 6 are trivial. So if the order of the group is 5 then the order of the subgroups are 1 and 5 which are trivial Only subgroups are {0}, G, {0,4,6} and {0,2}. Ex 10C pg. 402 No.s 1, 2, 3 Ex 10 Misc pg. 406 No.s 6, 7, 8