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Given a subgroup \(H\) of a group \(G\), for \(a \in G\), the set
\[aH = \{ah: h \in H\}\]is called the left coset of \(H\) containing \(a\), and \(Ha\) is called the right coset of \(H\) containing \(a\). The set of left cosets of \(H\) in \(G\) is denoted by \(G/H\), and the set of right cosets of \(H\) in \(G\) is denoted by \(H\backslash G\).
Right coset 的 notation \(H\backslash G\) 怎麼看？和「\(G\) 除以 \(H\)（\(G/H\) ）」的 quotient 想法相同（注意除號的方向），只是把 \(G\) 擺在右邊而已！
For two left cosets \(aH\) and \(bH\), the following hold:
由於 coset 的 disjoint 性質，coset 其實隱含了一 equivalence relation：
\[g_1 \sim g_2, \text{ if } g_1 \text{ and } g_2 \text{ are in the same left coset.}\]Proof
1 . By Cayley’s Theorem we konw \(\lambda_a\) is a bijection. Since \(aH = \lambda_a(H)\),
2 . Suppose \(ah_1 = bh_2\) for some \(h1,h_2 \in H\). Since \(hH = H\) for all \(h \in H\), \(H\) and \(aH\) have the same cardinality.
\[aH = ah_1H = bh_2H = bH.\]3 . Suppose \(aH = bH\), we have \(ah=be=b\) for some \(h \in H\). … Conversely, if \(a^{-1}b=h\) for some \(h \in H\). …
4 . For all \(g \in G\), we have \(g \in gH\). Therefore, every element is contained in some left cosets. Together with (2), this can be proved.
(2) implies disjoint.
right coset 的 properties 自行想像。
Let \(G=\mathbb{Z}_{12}\) and \(H = \{\bar 0, \bar 3\}\). We have
\[\begin{align*} G &= \{\bar 0, \bar 3\} \sqcup \{\bar 1, \bar 4, \bar 7, \bar 10\} \sqcup \{\bar 2, \bar 5, \bar 8, \bar 11\} \\ &= H \sqcup (\bar 1 + H) \sqcup (\bar 2 + H). \end{align*}\]因為 \(+\) 可交換，所以 left coset 和 right coset 相同。
要找出所有 left coset，首先寫出 \(H\)，再找出一個 \(g\) 使得 \(g\in G\) 但 \(g \not \in H\)，然後乘上 \(H\)；重複直到全部找出來。