A subgroup of a group is a (nonempty) subset which is closed under composition () and inversion ().
Given two groups and , a map is a homomorphism if ,.
Kernel and image
Definition 3.1 (Kernel). The kernel of a group homomorphism is
Proposition 3.2.
is a subgroup of (check that it contains , products, inverses).
is injective iff (using: ).
Definition 3.3 (Image). The image of a group homomorphism is
This is a subgroup of . is surjective iff .
Remark. If is injective, then is isomorphic to the subgroup . (The isomorphism is given by the map ,.)
Subgroups of
Given , is a subgroup.
Proposition 3.4. All nontrivial subgroups of are of this form.
Proof. This follows from the Euclidean algorithm. Given a nontrivial subgroup , there exists such that . Let be the smallest positive element of . Given any , for some and (the remainder). Since and ,. Since , by definition of , must be zero. Hence ; so , and conversely , so .∎
So every subgroup of is generated by a single element , in the following sense.
Question 3.5. Given a (nonempty) subset , what is the smallest subgroup of which contains ? This is denoted and called the subgroup generated by .
Answer: look at all subgroups of which contain (there’s at least itself!) and take their intersection:
More useful answer: must contain all products of elements of and their inverses, and these form a subgroup of , so
Definition 3.6. A group is cyclic if it is generated by a single element.
(Examples: ,. These are in fact the only cyclic groups up to isomorphism.)
Example 3.7. Let be any element in a group . Then the map , is a homomorphism, with image , the cyclic subgroup generated by .
Definition 3.8. The order of is the smallest positive such that , if it exists. Else, say has infinite order.
Note. Do not confuse the order of with the order of (). Though, .
If has infinite order, then the powers of are all distinct, is injective, and is isomorphic to . If has finite order , then , and is isomorphic to . (This completes the classification of cyclic groups, by the way.)
Example 3.9.
(observe: has order 6, so it generates).
Similarly, . But : for all , versus .
Classifying finite groups
An important question in group theory is the classification of finite groups up to isomorphism. This becomes increasingly difficult as increases. The beginning:
every group of order 2 is isomorphic to (by writing the table of the composition law…);
similarly, every group of order 3 is ;
for order 4, we know and . (These are different: every nonzero element of has order 2, while has an element of order 4.) In fact these are the only two groups of order 4 up to isomorphism.
(The classification of finite simple groups was completed in the 1980s, taking thousands of pages. We’ll learn some of the key tools and concepts in the class, but certainly won’t tackle the complete classification!)
Aside: equivalence relations and partitions
(Cf. Artin §2.7; also Halmos, Naive Set Theory.)
An equivalence relation on a set is a way to declare certain elements equivalent to each other (“”), yielding a smaller set of equivalence classes(“”, the quotient of by ).
Definition 3.10. An equivalence relation on a set is a binary relation (i.e. a subset of ; write iff is in this subset) which is
reflexive: ,;
symmetric: ,;
transitive: , if and then .
The equivalence class of is (sometimes denoted ). (By transitivity, the elements of are all equivalent to each other.)
The equivalence classes form a partition of , i.e. they are mutually disjoint subsets of whose union is .
The quotient of by is the set of equivalence classes: . This comes with a surjective map
Example 3.11 (Congruence mod ).; given , set iff divides . This is congruence mod ; check it is an equivalence relation.
There are equivalence classes:
The quotient is naturally in bijection with :
(We defined as only to avoid the language of equivalence classes, but it makes more sense to redefine it as the quotient set.)
Example 3.12 (Fibres of a surjection). Given a surjective , set iff . This is an equivalence relation; the partition into equivalence classes is
Note that is in bijection with via .
Using this:
More generally, given a map and an equivalence relation on , when does induce a map defined by ?
Necessary and sufficient: is compatible with , in the sense that .
This is exactly what ensures that, given an equivalence class in , its proposed image under doesn’t depend on the choice of element in it! We can then factor
Similarly, the reason that the operation induces an operation on is that it is compatible with :( divides and divides ). So we can define .
Notes not reached in this lecture, covered at the start of {{LINK:L04}}
Back to groups: cosets
Assume we have a surjective group homomorphism . Recall the kernel is a subgroup of .
Let’s look at the partition of induced by :
(for the last step, let ; then ).
Definition 3.13 (Cosets). Given any subgroup of a group :
is called the (left) coset of containing .
The relation is an equivalence relation on , whose equivalence classes are the left cosets.
The quotient (the set of left cosets) is denoted by . We have a partition
Proof.
, so for all .
If , then , hence , hence .
If and , then and , so , i.e. .
Also, such that such that .∎
Example 3.14., has kernel : the cosets are (), and we have a bijection
This gives a group law on the quotient! (Addition of cosets addition mod .)
When a subgroup is the kernel of a homomorphism , we get a bijection
(recall iff ), and we can use this bijection to get a group structure on , essentially (which corresponds via to ). Then
is a group homomorphism.
For a general subgroup , however, trying to make a group by setting might not work. The obstacle to this is: assume () and (). Does it follow that ()? If not, our operation isn’t well defined.
Example 3.15. = symmetries of the square, where = horizontal flip. Then (the coset ). But setting = rotation by , is the reflection across one diagonal, while the coset of is , where is the reflection across the other diagonal.