Conditions for a subset to be a subgroup

Dependencies:

1. Group
2. Subgroup
3. Identity of a group is unique
4. Inverse of a group element is unique

Let $H$ be a subset of $G$. $H$ is a subgroup of $G$ iff

• $H$ is closed under $*$.
• $\operatorname{id}(G) \in H$.
• $\forall h \in H, h^{-1} \in H$.

Proof

• Closure: Given.

• Associativity: True since $H$ inherits its operator from $G$.

• Identity: Let $e_G = \operatorname{id}(G)$ and $e_H = \operatorname{id}(H)$ (if it exists). We'll prove both sides of the implication:

  • $\forall h \in H, he_G = e_Gh = h$ because $h \in G$. Since identity of a group is unique, $e_G \in H \Rightarrow e_H \textrm{ exists}$.
  • Now assume that $e_H$ exists. $e_He_H = e_H$ because $e_H = \operatorname{id}(H)$. $e_He_G = e_H$ because $e_G = \operatorname{id}(G)$ and $e_H \in G$. Therefore, $e_G = e_H \Rightarrow e_G \in H$.

• Inverse: Since inverse is unique in $G$, only the inverse in $G$ has the possibility of being an inverse in $H$.

Dependency for:

1. Condition for a subset to be a subgroup
2. Cyclic Group

Info:

• Depth: 2
• Number of transitive dependencies: 4

Transitive dependencies:

1. Group
2. Identity of a group is unique
3. Subgroup
4. Inverse of a group element is unique