Condition for a subset to be a subgroup
Dependencies:
$H$ is a subgroup of $G$ iff $H \neq \phi$ and $(h_1, h_2 \in H \Rightarrow h_1 h_2^{-1} \in H)$.
Proof
We will prove both sides of the implication.
Part 1
Let $H$ be a group. Then $H \neq \phi$.
$h_1, h_2 \in H \Rightarrow h_1, h_2^{-1} \in H \Rightarrow h_1 h_2^{-1} \in H$.
Therefore, if $H$ is a subgroup of $G$, then $H \neq \phi$ and $(h_1, h_2 \in H \Rightarrow h_1 h_2^{-1} \in H)$.
Part 2
$H \neq \phi \Rightarrow \exists h \in H$.
In $(h_1, h_2 \in H \Rightarrow h_1 h_2^{-1} \in H)$:
- Substitute $(h_1=h, h_2=h)$ to get $\operatorname{id}(G) \in H$.
- Substitute $(h_1=\operatorname{id}(G), h_2=h)$ to get $h^{-1} \in H$.
- $(h_1=h_1, h_2=h_2^{-1})$ to get $h_1 h_2 \in H$.
By the above 3 properties, $H$ is a subgroup of $G$.
Dependency for:
- Second isomorphism theorem
- Alternating Group
- Correspondence theorem
- Homomorphic mapping of subgroup of domain is subgroup of codomain
- Conditions for a subset of a ring to be a subring
- Condition for being a subspace
- F[x]/p(x): A ring
Info:
- Depth: 3
- Number of transitive dependencies: 5