## An arbitrary intersection of subgroups is a subgroup

Let $G$ be a group.

1. Prove that if $H$ and $K$ are subgroups of $G$, then so is $H \cap K$.
2. Prove that if $\{H_i\}_{i \in I}$ is a family of subgroups of $G$ then so is $\bigcap_{i \in I} H_i$.

1. Note that $H \cap K$ is not empty since $1 \in H \cap K$. Now suppose $x, y \in H \cap K$. Then since $H$ and $K$ are subgroups, we have $xy^{-1} \in H$ and $xy^{-1} \in K$ by the subgroup criterion; thus $xy^{-1} \in H \cap K$. By the subgroup criterion, $H \cap K$ is a subgroup of $G$.
2. Note that $\bigcap_{i \in I} H_i$ is not empty since $1 \in H_i$ for each $i \in I$. Now let $x,y \in \bigcap_{i \in I} H_i$. Then $x,y \in H_i$ for each $i \in I$, and by the subgroup criterion, $xy^{-1} \in H_i$ for each $i \in I$. Thus $xy^{-1} \in \bigcap_{i \in I} H_i$. By the subgroup criterion, $\bigcap_{i \in I} H_i$ is a subgroup of $G$.
Advertisements
Post a comment or leave a trackback: Trackback URL.