Exercise 2.1.10 (Intersection of subgroups)

Proof.

(a)

Suppose that $H$ and $K$ are subgroups of $G$. Then

• $1_G\in H$ and $1_G\in K$, thus $1_G\in H\cap K$ and $H\cap K\ne \varnothing$.
• If $x\in H\cap K$ and $y\in H\cap K$, then $x\in H$ and $y\in H$, thus $x{y}^{-1}\in H$. Similarly $x\in K$ and $y\in K$, thus $x{y}^{-1}\in K$. Therefore $x{y}^{-1}\in H\cap K$.

If $H$ and $K$ are subgroups of $G$ then so is their intersection $H\cap K$.

(b)

Consider a family ${(H_i)}_{i\in I}$ of subgroups of $G$ (where $I$ may have any cardinality). Put $H=\underset{i\in I}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}{H}_i$.

• For all $i\in I$, $1_G\in H_i$, thus $1_G\in \underset{i\in I}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}{H}_i$, so $H\ne \varnothing$.
• Suppose that $x\in H$ and $y\in H$. Then $x\in H_i$ and $y\in H_i$ for all $i\in I$. Since every $H_i$ is a subgroup of $G$, $x{y}^{-1}\in H_i$ for all $i\in I$, therefore $x{y}^{-1}\in \underset{i\in I}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}{H}_i=H$.

If $H_i$ is a subgroup of $G$ for all $i\in I$, then $\underset{i\in I}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}{H}_i$ is a subgroup of $G$.