GH = H Iff G In H - TheoremDep

gH = H iff g in H

Dependencies:

1. Coset
2. Inverse of a group element is unique

Let $H$ be a subgroup of $G$ and $g \in G$. Then $gH = H \iff g \in H$.

Proof

$g \in H \Rightarrow gH \subseteq H$, because of closure in $H$.

Since inverse of $g$ is unique and $H$ is a group, $g^{-1} \in H$. Therefore, $\forall h \in H, g^{-1}h \in H$ by closure of $H$.

$h \in H \Rightarrow h = g(g^{-1}h) \in gH \Rightarrow H \subseteq gH$. Therefore, $H = gH$.

Let $H$ be a subset of $G$ such that $gH = H$. \begin{align} gH = H &\Rightarrow gH \subseteq H \\ &\Rightarrow \forall h_1 \in H, \exists h_2 \in H, gh_1 = h_2 \\ &\Rightarrow \forall h_1 \in H, \exists h_2 \in H, g = h_2h_1^{-1} \\ &\Rightarrow g \in H \end{align}

Dependency for:

1. Third isomorphism theorem
2. Second isomorphism theorem
3. First isomorphism theorem
4. I is a prime ideal iff R/I is an integral domain
5. I is a maximal ideal iff R/I is a field
6. Two cosets are either identical or disjoint
7. Product of ideal cosets is well-defined
8. F[x]/p(x) is isomorphic to F[x]/p(x)F[x]

Info:

• Depth: 2
• Number of transitive dependencies: 3

Transitive dependencies:

1. Group
2. Coset
3. Inverse of a group element is unique