2 - Subgroups

Subgroup

let $G$ be a group, then a subset $H$ of $G$ is a subgroup of $G$ if it a group, inheriting the operator from $G$ . In such a case we write $H ⩽ G$ . if $H \neq G$ , we write $H < G$ .

necessary and sufficient conditions for subgroups

let $G$ be a group, $H$ , a subset of $G$ is a subgroup if and only if the following properties hold:

The identity $e \in G$ is also in $H$ .
$x, y \in H$ implies that $x y \in H$ ( $H$ is closed under the product operator inherited from $G$ ).
$x \in H$ implies that $x^{- 1} \in H$ ( $H$ is closed under inverses).

\begin{proof} if $H$ is a subgroup, 1,2,3 follow by definition. Now suppose $H \subset G$ . Then the above three conditions guarantee that $H$ is a group, with associativity following from inheriting the operator from $G$ .

\end{proof}

A simpler condition to guarantee subgroups

let $G$ be a group, and $H$ a subset of $G$ . then $H$ is a subgroup if and only if $H$ is non empty, and $x, y \in H$ implies $x y^{- 1} \in H$ .

\begin{proof} Suppose $H$ was a subgroup of $G$ . then obviously, $H$ is non empty (it at least contains $e$ , the identity) and $x, y \in H$ implies $x y^{- 1} \in H$ . Now, suppose $H$ was a non empty subset of $G$ , and that $x, y \in H$ implies $x y^{- 1} \in H$ . We will show that $e \in H$ : notice that $x \in H$ implies $x x^{- 1} = e \in H$ . Now we show that $H$ is closed under inverses: Since $e \in H$ , for any element $x \in H$ , $e x^{- 1} = x^{- 1} \in H$ . Finally, we will show that $H$ is closed under products: Suppose $x, y \in H$ . Then $y^{- 1} \in H$ , (as shown earlier). Hence $x (y^{- 1})^{- 1} \in H$ . Hence $x y \in H$ . Therefore by [[#^97f5ee]] $H$ is a subgroup of $G$ .

\end{proof}

The intersection of subgroups

Let $G$ be a group, and $H, K$ be subgroups of $G$ . Then, $H \cap K$ is a subgroup of $G$ .

\begin{proof} since $e$ is in both $H$ and $K$ , $H \cap K$ is non empty. Suppose $x, y \in H \cap K$ . Then $x, y$ is in both $H$ and $K$ . Therefore, $x y^{- 1}$ is in both $H$ and $K$ . Hence, $x y^{- 1} \in H \cap K$ . See :[[#^5c6133]].

\end{proof}

The intersection of finitely many Subgroups is a subgroup

Let $H_{1}, H_{2}, \dots H_{n}$ All be subgroups of a group $G$ . Then, $⋂_{i = 1}^{n} H_{i}$ is a subgroup of $G$ .

Proof Outline

Use [[#^c26bc4]] with a natural induction argument.

Finiteness with product closure are sufficient conditions for a subgroup

Let $H$ be finite subset of a group $G$ , such that the identity $e \in H$ . If $x, y \in H$ implies $x y \in H$ , then $H$ is a subgroup of $G$ .

An intuition for the above idea:

![[diagram-20240712 (2).svg#invert_B]]
Imagine a finite set $H$ , closed under products, with the identity element $e$ in it. Starting from an arbitrary element $x$ , start taking left products, $h_{1} x$ , $h_{2} h_{1} x$ and so on... each of these products are the images of $x$ under distinct left coset maps. so eventually, we have to hit $e$ , (we will show that we can't loop around without ever hitting $e$ ), basically allowing us to get closure under inverses.

\begin{proof}
Since $H$ is finite, let $| H | = n$ , we can index the elements of $H = {h_{0} = e, h_{1}, \dots h_{n - 1}}$ . We claim that the left coset maps $L (H)$ are well defined from $H$ to $H$ . suppose $x \in H$ and $h_{i} \in H$ . then $f_{h_{i}} (x) = h_{i} x \in H$ due to product closure.
Now, Consider the set of orbits of $x \in H$ under each map in $L (H)$ defined as $O r b (x) = {h_{i} x : h_{i} \in H}$ . By two left coset maps never agree on the image of a particular element if $h_{i} x = h_{j} x$ , then in $G$ which has inverses, $h_{i} = h_{j}$ meaning that they are the identical left coset map in $G$ , Hence, they are identical in $H$ . Therefore $O r b (x)$ has $n$ distinct elements, each of which lie in $H$ . Hence $O r b (x) = H$ , given that $H$ is finite. Since $O r b (x) = H$ there exists some $h_{j} \in H$ such that $h_{j} x = h_{0} = e$ . This is the inverse of an element $x$ . Hence $H$ is closed under inverses. see [[#^97f5ee]].

\end{proof}