# 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 $(G, \cdot)$ be a group. If $S \subseteq G$ and $S$ forms a group under the same operation $\cdot$ then $(S, \cdot)$ is said to be a Subgroup of $(G, \cdot)$. If $S$ is a subgroup of $G$ then we write $S \leq G$. |

Definition: Let $(G, \cdot)$ be a group. If $(H, \cdot)$ is a group such that $G \subseteq H$ then $(H, \cdot)$ is said to be a Group Extension of $(G, \cdot)$. |

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 $(S, \cdot)$ is a subgroup of $(G, \cdot)$ where $S \subseteq G$, that then we only need to check two of the four group axioms for verification

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

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

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

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

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

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

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

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

**Proof:**$\Rightarrow$ If $(S, \cdot) $] is a subgroup of [[$ (G, \cdot)$ then this direction is trivial.

- $\Leftarrow$ Suppose that $S \neq \emptyset$ and for all $a, b \in S$ we have that $a \cdot b^{-1} \in S$. Since $S \neq \emptyset$ there exists an $a \in S$. So $a \cdot a^{-1} = e \in S$.

- Now if $a \in S$, then since $e \in S$ we have that $e \cdot a^{-1} = a^{-1} \in S$. So if $a \in S$ then $a^{-1} \in H$.

- Lastly, if $a, b \in S$ then $a, b^{-1} \in S$. Thus $a \cdot (b^{-1})^{-1} = a \cdot b \in S$. So $S$ is closed under the operation $\cdot$. Thus $(S, \cdot)$ is a subgroup of $(G, \cdot)$. $\blacksquare$