# Changes

## Left cosets partition a group

, 00:07, 4 July 2011
Proof in form (4)
==Equivalence of statements==
These statements are equivalent because of the following general fact about sets and equivalence relations. If $S$ is a set, and $\sim$ is an equivalence relation on $S$, then we can partition $S$ as a disjoint union of ''equivalence classes'' under $\! \sim$. Two elements $a$ and $b$ are defined to be in the same equivalence class under $\! \sim$ if $\! a \sim b$.
Conversely, if $S$ is partitioned as a disjoint union of subsets, then the relation of being in the same subset is an equivalence relation on $S$.
Let $G$ be a group, $H$ be a subgroup.
For $a,b \in G$, we say that $a$ is in the left coset of $b$ with respect to $H$ if there exists $h \in H$ such that $a = bh$.
<section end=beginner/>
<section begin=revisit/>
===Reflexivity===
'''To prove''': For any $a \in G$, $\! a \sim a$.
'''Proof''': Clearly $e \in H$ (since $H$ is a subgroup). Hence, for any $a \in G$, $a = ae$, so $\! a \sim a$: $a$ is in its own left coset.
===Symmetry===
'''To prove''': For any $a,b \in G$ such that $\! a \sim b$, we have $\! b \sim a$.
'''Proof''': 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 $a$. In symbols, $a \sim b \implies b \sim a$.
===Transitivity===
'''To prove''': If $a,b,c \in G$ are such that $\! a \sim b$, and $\! b \sim c$, then $a \sim c$
'''Proof''': 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 $c$.
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 $bh = ah_1h_2^{-21}h \in aH$. Thus, $aH \subset subseteq bH$ and $bH \subset subseteq aH$, so $aH = bH$.
<section end=beginner/>
<section begin=revisit/>

==Other proofs==