# Subgroups and Group Extensions

We have just looked at the definition of a group on the Groups page. Before we move on though, we will need to look at a couple of other key definitions regarding groups - namely groups that are somewhat contained within another.

Definition: Let $(G, *)$ be a group. If $S \subseteq G$ and $S$ forms a group under the same operation $*$ then $(S, *)$ is said to be a Subgroup of $(G, *)$. Furthermore, $(G, *)$ is said to be a Group Extension of $(S, *)$. |

*The notation "$S \leq G$" is used to denote that $S$ is a subgroup of $G$.*

For example, consider the group of complex numbers under the operation of standard addition, $(\mathbb{C}, +)$. We know that the set of real numbers is a subset of the set of complex numbers, that is, $\mathbb{R} \subset \mathbb{C}$ and so the group of real numbers under the operation of standard addition, $(\mathbb{R}, +)$ is a subgroup of $(\mathbb{C}, +)$ and $(\mathbb{C}, +)$ is a group extension of $(\mathbb{R}, +)$.

We will now look at a nice theorem which tells us that to determine if $(S, *)$ is a subgroup of $(G, *)$ where $S \subseteq G$, that then we only need to check two of the four group axioms for verification

Theorem 1: If $(G, *)$ is a group with the identity $e \in G$ of $*$ and $S \subseteq G$ then $(S, *)$ is a subgroup of $(G, *)$ if and only if $S$ is closed under $*$ and for all $a \in S$ we have that $a^{-1} \in S$ such that $a * a^{-1} = e$ and $a^{-1} * a = e$. |

**Proof:**Let $(G, *)$ be a group with the identity $e \in G$ of $*$ and let $S \subseteq G$.

- $\Rightarrow$ Suppose that $(S, *)$ is a subgroup of $(G, *)$. Then by definition, $(S, *)$ is a group itself and satisfies all of the group axioms - namely that $S$ is closed under the operation $*$ and that for all $a \in S$ there exists an $a^{-1} \in S$ such that $a * a^{-1} = e$ and $a^{-1} * a = e$.

- $\Leftarrow$ Now suppose that $S$ is closed under $*$ and that for all $a \in S$ there exists an $a^{-1} \in S$ such that $a * a^{-1} = e$ and $a^{-1} * a = e$. These are precisely two of the group axioms we have looked at, and to show that $(S, *)$ is a subgroup of $(G, *)$ we only need to show that the other two axioms hold.

- First suppose that $a, b, c \in S$ and that $a * (b * c) \neq (a * b) * c$, that is, suppose that $*$ is not associative on $S$. Since $S \subseteq G$ we must have that $a * (b * c) \neq (a * b) * c$ for this particular $a, b, c \in G$ which contradicts the associativity of $*$ on the group $G$. Hence $*$ must actually be associative on $G$.

- Now since $S$ is closed under $*$ and for $a \in S$ there exists an $a^{-1} \in S$ such that $a * a^{-1} = e$ and $a^{-1} * a = e$ we must have that $e \in S$ and furthermore, $a * e = a$ and $e * a = a$.

- Therefore $(S, *)$ is a group, and in particular since $S \subseteq G$ we have that $(S, *)$ is a subgroup of $(G, *)$. $\blacksquare$

Theorem 2: If $(G, *)$ is a group with the identity $e \in G$ of $*$ and $S \subseteq G$ then $(S, *)$ is a subgroup of $(G, *)$ if and only if $H \neq \emptyset$ and for all $a, b \in H$ we have that $a * b^{-1} \in H$. |