Difference between revisions of "Left cosets partition a group"
(→Equivalence of statements) |
(→Statement with symbols) |
||
Line 18: | Line 18: | ||
# The left cosets of <math>H</math>, namely <math>gH, g \in G</math>, form a partition of the group <math>G</math>. In other words, <math>G</math> is a disjoint union of left cosets of <math>H</math>. | # The left cosets of <math>H</math>, namely <math>gH, g \in G</math>, form a partition of the group <math>G</math>. In other words, <math>G</math> is a disjoint union of left cosets of <math>H</math>. | ||
# The relation <math>a \sim b \iff a \in bH</math> is an equivalence relation on <math>G</math> | # The relation <math>a \sim b \iff a \in bH</math> is an equivalence relation on <math>G</math> | ||
− | # For every <math>g \in G</math>, there is ''exactly'' one left coset containing <math>g</math>. | + | # For every <math>g \in G</math>, there is ''exactly'' one left coset of <math>H</math> in <math>G</math> containing <math>g</math>. |
# If <math>aH</math> and <math>bH</math> are left cosets of <math>H</math> in <math>G</math>, then either <math>aH = bH</math> or <math>aH \cap bH</math> is empty. | # If <math>aH</math> and <math>bH</math> are left cosets of <math>H</math> in <math>G</math>, then either <math>aH = bH</math> or <math>aH \cap bH</math> is empty. | ||
Revision as of 21:38, 4 May 2010
DIRECT: The fact or result stated in this article has a trivial/direct/straightforward proof provided we use the correct definitions of the terms involved
View other results with direct proofs
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this
Statement
Verbal statement
The following equivalent statements are true:
- The left cosets of a subgroup in a group partition the group.
- The relation of one element being in the left coset of the other, is an equivalence relation.
- Every element of the group is in exactly one left coset.
- Any two left cosets of a subgroup either do not intersect, or are equal.
Statement with symbols
Suppose is a group, and is a subgroup. Then, the following equivalent statements are true:
- The left cosets of , namely , form a partition of the group . In other words, is a disjoint union of left cosets of .
- The relation is an equivalence relation on
- For every , there is exactly one left coset of in containing .
- If and are left cosets of in , then either or is empty.
Equivalence of statements
These statements are equivalent because of the following general fact about sets and equivalence relations. If is a set, and is an equivalence relation on , then we can partition as a disjoint union of equivalence classes under . Two elements and are defined to be in the same equivalence class under if .
Conversely, if is partitioned as a disjoint union of subsets, then the relation of being in the same subset is an equivalence relation on .
Hence, there is a correspondence between equivalence relations on a set and partitions of the set into subsets. This statement about left cosets thus states that the left cosets partition the group, which is also the same as saying that the relation of one element being in the left coset of another, is an equivalence relation.
Here, we give the proof both in form (2) and form (4). The two proofs are essentially the same, but they are worked out in somewhat different language, and explain how to think both in terms of equivalence relations and in terms of partitions.
Definitions used
Let be a group, be a subgroup.
For , we say that is in the left coset of if there exists such that .
Related facts
Converse
A partial converse to this result is true. If is a subset of containing the identity element with the property that the set of all left translates of , i.e. the set of subsets , form a partition of , then is a subgroup of .
Further information: Subset containing identity whose left translates partition the group is a subgroup
Analogues in other algebraic structures
The proof that the left cosets of a subgroup partition the group uses all the properties of groups: the existence of identity element is used to prove reflexivity, the existence of inverses (along with associativity and the identity element) was used to prove symmetry, and associativity is used to prove transitivity. Hence, extending the result to algebraic structures weaker than groups is in general hard. There are, however, some ways of extending.
Statement | Analogue of group | Analogue of subgroup | Comment |
---|---|---|---|
Left cosets of a subgroup partition a monoid | monoid (associative, identity element, not necessarily any inverses) | submonoid that is in fact a group | We do not require the bigger structure to be a group. All we need is associativity in the bigger structure. Thus, the left cosets of a subgroup in a monoid, still partition it. (Note that we still do require associativity in the bigger structure). |
Relation with right cosets and normal subgroups
- Right cosets partition a group: The proof of this is analogous to that for left cosets.
- Left cosets are in bijection via left multiplication: In particular, any two left cosets of a subgroup have the same size as the subgroup.
- Equivalence of definitions of coset: A subset of a group occurs as the left coset of a subgroup if and only if it occurs as the right coset of a subgroup.
- For a normal subgroup, the set of left cosets coincides with the set of right cosets, and this set can be given the structure of a group called the quotient group. Any homomorphism of groups is obtained by composing the quotient map to a quotient group with an injection: see normal subgroup equals kernel of homomorphism and first isomorphism theorem.
- Lattice of subgroups embeds in partition lattice: As we see here, every subgroup gives rise to a partition of the group (namely, the partition into left cosets). This gives a function from the lattice of subgroups of a group to the partition lattice of the group. It turns out that this map is a lattice embedding, i.e., it preserves the lattice operations.
Proof in form (2)
Given: A group , a subgroup
To prove: The relation such that , is an equivalence relation on
Reflexivity
To prove: For any , .
Proof: Clearly (since is a subgroup). Hence, for any , , so : is in its own left coset.
Symmetry
To prove: For any such that , we have .
Proof: If , for some , then . Since and is a subgroup, . Thus, if is in the left coset of , then is in the left coset of . In symbols, .
Transitivity
To prove: If are such that , and , then
Proof: If , and , for , and . Since is a subgroup, , so is in the left coset of .
Proof in form (4)
Given: A group , a subgroup , two elements
To prove: The left cosets and are either equal or disjoint (they have empty intersection)
Proof: We'll assume that and are not disjoint, and prove that they are equal.
For this, suppose . Then, there exist such that . Thus, and .
Now, for any element , we have , and similarly, for every element , we have . Thus, and , so .
Other proofs
Orbits under a group action
One easy way of seeing that the left cosets partition a group is by viewing the left cosets as orbits of the group under the action of the subgroup by right multiplication.
Left congruence
Another way of viewing the partition of a group into left cosets of a subgroup is in terms of a left congruence. A left congruence on a magma is an equivalence relation with the property that:
The only left congruences on a group are those that arise as partitions in terms of left cosets of a subgroup.
References
Textbook references
- Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Proposition 4, Page 80