# Difference between revisions of "Left cosets partition a group"

Jump to: navigation, search

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:

1. The left cosets of a subgroup in a group partition the group.
2. The relation of being in the same left coset is an equivalence relation.
3. Every element of the group is in exactly one left coset.
4. Any two left cosets of a subgroup either do not intersect, or are equal.

## Definitions used

Let $G$ be a group, $H$ be a subgroup.

For $a,b \in G$, we say that $a$ is in the left coset of $b$ if there exists $h \in H$ such that $a = bh$.

## Proof in form (2)

The various statements are clearly equivalent. We'll prove (2) here: the relation of being in the same left coset is an equivalence relation.

### Reflexivity

Clearly $e \in H$ (since $H$ is a subgroup). Hence, for any $a \in G$, $a = ae$, so $a$ is in the left coset of $a$.

### Symmetry

If $a = bh$, for some $h \in H$, then $b = ah^{-1}$. Since $h \in H$ and $H$ is a subgroup, $h^{-1} \in H$. Thus, if $a$ is in the left coset of $b$, then $b$ is in the left coset of $a$.

### Transitivity

If $a = bh$, and $b = ck$, for $h, k \in H$, and $a = ckh$. Since $H$ is a subgroup, $h,k \in H \implies kh \in H$, so $a$ is in the left coset of $c$.

## Proof in form (4)

Here, we'll show that given two left cosets $aH$ and $bH$ of a subgroup $H$, either $aH = bH$ or $aH \cap bH$ is empty.

For this, suppose $c \in aH \cap bH$. Then, there exist $h_1,h_2$ such that $ah_1 = bh_2 = c$. Thus, $b = ah_1h_2^{-1} \in aH$ and $a = bh_2h_1^{-1} \in bH$.

Now, for any element $ah \in aH$, we have $ah = bh_2h_1^{-1}h \in bH$, and similarly, for every element $bh \in bH$, we have $bh = ah_1h_2^{-2}h \in aH$. Thus, $aH \subset bH$ and $bH \subset aH$, so $aH = bH$.

## 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 $(S,*)$ is an equivalence relation $\sim$ with the property that:

$a \sim b \implies c * a \sim c * b \ \forall \ c \in S$

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