Jump to: navigation, search

Left cosets partition a group

57 bytes added, 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 <math>S</math> is a set, and <math>\sim</math> is an equivalence relation on <math>S</math>, then we can partition <math>S</math> as a disjoint union of ''equivalence classes'' under <math>\! \sim</math>. Two elements <math>a</math> and <math>b</math> are defined to be in the same equivalence class under <math>\! \sim</math> if <math>\! a \sim b</math>.
Conversely, if <math>S</math> is partitioned as a disjoint union of subsets, then the relation of being in the same subset is an equivalence relation on <math>S</math>.
Let <math>G</math> be a group, <math>H</math> be a subgroup.
For <math>a,b \in G</math>, we say that <math>a</math> is in the left coset of <math>b</math> with respect to <math>H</math> if there exists <math>h \in H</math> such that <math>a = bh</math>.
<section end=beginner/>
<section begin=revisit/>
'''To prove''': For any <math>a \in G</math>, <math>\! a \sim a</math>.
'''Proof''': Clearly <math>e \in H</math> (since <math>H</math> is a subgroup). Hence, for any <math>a \in G</math>, <math>a = ae</math>, so <math>\! a \sim a</math>: <math>a</math> is in its own left coset.
'''To prove''': For any <math>a,b \in G</math> such that <math>\! a \sim b</math>, we have <math>\! b \sim a</math>.
'''Proof''': If <math>a = bh</math>, for some <math>h \in H</math>, then <math>b = ah^{-1}</math>. Since <math>h \in H</math> and <math>H</math> is a subgroup, <math>h^{-1} \in H</math>. Thus, if <math>a</math> is in the left coset of <math>b</math>, then <math>b</math> is in the left coset of <math>a</math>. In symbols, <math>a \sim b \implies b \sim a</math>.
'''To prove''': If <math>a,b,c \in G</math> are such that <math>\! a \sim b</math>, and <math>\! b \sim c</math>, then <math>a \sim c</math>
'''Proof''': If <math>a = bh</math>, and <math>b = ck</math>, for <math>h, k \in H</math>, and <math>a = ckh</math>. Since <math>H</math> is a subgroup, <math>h,k \in H \implies kh \in H</math>, so <math>a</math> is in the left coset of <math>c</math>.
For this, suppose <math>c \in aH \cap bH</math>. Then, there exist <math>h_1,h_2</math> such that <math>ah_1 = bh_2 = c</math>. Thus, <math>b = ah_1h_2^{-1} \in aH</math> and <math>a = bh_2h_1^{-1} \in bH</math>.
Now, for any element <math>ah \in aH</math>, we have <math>ah = bh_2h_1^{-1}h \in bH</math>, and similarly, for every element <math>bh \in bH</math>, we have <math>bh = ah_1h_2^{-21}h \in aH</math>. Thus, <math>aH \subset subseteq bH</math> and <math>bH \subset subseteq aH</math>, so <math>aH = bH</math>.
<section end=beginner/>
<section begin=revisit/>
==Other proofs==
Bureaucrats, emailconfirmed, Administrators

Navigation menu