# Subgroups

Learn about subgroups and the order of groups in this lesson.

We'll cover the following

Definition:

Let $(G, *)$ be a group and $H \subset G$. $H$ is called a subgroup of $G$ (usually denoted by $H \leq G$ ) if $H$ also forms a group with respect to the restriction of the same binary operation $*$, i.e.,

H1 for all $a, b \in H \Rightarrow a * b \in H$ ($H$ is closed with respect to $*) .$

H2 $H$ together with the induced binary operation.

$*: H \times H \rightarrow H,(a, b) \mapsto a * b$

forms a group, meaning that the restriction of $*$ to $H \times H$ is a binary group operation on $\mathrm{H}$.

In order to form a subgroup of any group $G, H$ has to fulfill the following minimal requirements:

## Lemma 1: Subgroup criteria

Let $(G, *)$ be a group, $H \subset G$ with $H \neq \varnothing .$ Then, the following statements are equivalent:

1. $H \leqslant G$

2. $a, b \in H \Rightarrow a * b \in H$ and $a^{-1} \in H$

3. $a, b \in H \Rightarrow a * b^{-1} \in H$

Proof:

• (1) $\Rightarrow$ (2): Let $H \leqslant G$. According to the above definition, there are $a, b \in H$ and also $a * b \in H$. Since $H$ is a group, there exists an identity element $e^{\prime}$. Because $e^{\prime} * e^{\prime}=e^{\prime}$, as the concept discussed here :Corollary_2_6_1 yields $e^{\prime}=e$. Furthermore, since $H$ is a group, there exists an inverse $a^{\prime} \in H$ for every $a \in H$ such that $a^{\prime} * a=e$. Therefore, $a^{-1}=a^{\prime} \in H$.

• (2) $\Rightarrow$ (3): According to statement (2), if $a, b \in H$, then $a \in H$ and also $b^{-1} \in H$, therefore (2) yields $a * b^{-1} \in H$.

• (3) $\Rightarrow$ (1): For $a, b \in H$, (3) yields $b * b^{-1} \in H$, and thus $e=b * b^{-1} \in H$. It follows that $b^{-1}=e * b^{-1}$ for all $b \in H$, and therefore $a * b=a *\left(b^{-1}\right)^{-1}$. Altogether, $a, b \in H \Rightarrow a * b \in H$, which yields $H \leq G$ according to the definition given above (H1).

These minimal demands are defined in this lemma together with this lemma :Lemma_2_6_2 yield very natural conditions for an easy verifying of $H$ to be a subgroup of $G$.

## Theorem 1: subgroup verification

$H \leqslant G$ if, and only if, the following conditions are satisfied:

1. $H$ is closed with respect to the binary operator $*$, i.e., $a * b \in H$ for every $a, b \in H$.

2. $\mathrm{H}$ contains the identity element $e_{G}$ of $G$.

3. For each $a \in H, H$ also contains its inverse element, i.e., if $a, a^{-1} \in G$ and $a \in H$, then also $a^{-1} \in H$.

Note: For any group $G=(G, *)$ with identity $e_{G},\left\{e_{G}\right\}$ and $G$ itself are always subgroups of $G$, whereas $\left\{e_{G}\right\}$ is called the trivial subgroup. Any other subgroup is said to be a proper subgroup.

Example:

$(\mathbb{Z},+)$ is a subgroup of $(\mathbb{Q},+)$ and $(\mathbb{R},+)$. It’s easy to see that the set of integers $\mathbb{Z}$ is closed with respect to $+$, meaning that $a+b \in \mathbb{Z}$ for every $a, b \in \mathbb{Z}$. Furthermore, the identity element 0 of $\mathbb{Q}$ and $\mathbb{R}$ is also contained in $\mathbb{Z}$. Additionally, $\mathbb{Z}$ contains the inverse $-a$ for all $a \in \mathbb{Z}$.

Example:

$\left(\mathbb{Z}_{n},+\right)$ is a subgroup of $(\mathbb{Z},+)$.

Get hands-on with 1200+ tech skills courses.