For subset H of a group, a(bH) = (ab)H and (aH)b = a(Hb)

Dependencies:

1. Coset

Let $H$ be a subset of $G$. Let $a, b \in G$. Then $a(bH) = (ab)H$ and $(aH)b = a(Hb)$.

Proof

\begin{align} & a(bH) \\ &= a\{bh: h \in H\} \\ &= \{a(bh): h \in H\} \\ &= \{(ab)h: h \in H\} \\ &= (ab)H \end{align}

\begin{align} & (aH)b \\ &= \{ah: h \in H\}b \\ &= \{(ah)b: h \in H\} \\ &= \{a(hb): h \in H\} \\ &= a\{hb: h \in H\} \\ &= a(Hb) \end{align}

Dependency for:

1. Product of normal cosets is well-defined

Info:

• Depth: 2
• Number of transitive dependencies: 2

Transitive dependencies:

1. Group
2. Coset