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

$H \leqslant G$

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

$a, b \in H \Rightarrow a * b^{1} \in H$
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
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, $a^{1}=a^{\prime} \in H$.: Corollary_2_6_1 
(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
Theorem 1: subgroup verification
$H \leqslant G$ if, and only if, the following conditions are satisfied:

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

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

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 $(\mathbb{Z},+)$.
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