> [!NOTE] Theorem (Necessary condition for subgroup) > Let $(G,\cdot)$ be a [[Groups|group]]. Suppose $H\subset G$ then $(H,\cdot)$ is a [[Subgroup|subgroup]] iff > 1. Identity element: $1\in H$ > 2. Closure: $\forall a \forall b ( a,b\in H \implies a\cdot b\in H)$ > 3. Inverse: $\forall a (a\in H \implies a^{-1}\in H).$ ###### Proof ($\implies$) Suppose $(G,\cdot)$ is a group and $H\subset G$ that satisfies $(1),(2)$ and $(3).$ Closure: The closure condition is given by condition $(2).$ Associativity: For all $a,b,c\in H$ we have $a,b,c\in G$ and associativity holds in $G$ and so holds in $H.$ Identity: condition $(1)$ gives the existence of the identity element in $H.$ Inverse: finally condition $(3)$ asserts the existence of an inverse for every $a\in H.$ ($\impliedby$) Suppose $(H,\cdot)$ is a subgroup of $(G,\cdot).$ Let $1_{G}\in G$ and $1_{H}\in H$ denote the identities of $G$ and $H.$ Then we have $1_{H}=1_{H}1_{H}$Since $1_{H}\in G,$ it has a unique inverse $1_{H}^{-1}\in G.$ Multiplying both sides of the above equation gives $1^{-1}_{H}1_{H} = 1^{-1}_{H}1_{H} 1_{H} \implies 1_{G} = 1_{G} 1_{H} =1_{H}$and so $1_{G} \in H$ and condition $(1)$ is satisfied. Condition $(2)$ is satisfied since $H$ is group under the same binary operation as $G$ and so it is must be closed under this binary operation. Condition $(3)$ is satisfied since again $H$ is a group so every element has an inverse in $H.$ This inverse will the same in $G$ by [[Inverse of Group Element#^5bd480|uniqueness of inverses]]. # Application The [[Three Step Subring Test|three step subring test]] consists of the above conditions and additionally, closure of the set under the ring product.