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Problem 54 (The Intersection Of Two Subgroups)

Suppose that $G$ is a group and that $H$ and $K$ are subgroups of $G$. Prove that $H\cap K$ is a subgroup of $G$.

Solution

Suppose $G$ is a group and that $H$ and $K$ are subgroups of $G$. We will prove that $H\cap K$ is a subgroup of $G$ using the subgroup test.

First, it must be shown that $H\cap K$ is a nonempty subset of $G$. Since $G$ is a group and $H\leq G$ we know that the identity $e\in G$ implies $e\in H$. Similarly, $e\in K$. This means $e\in H\cap K$ and thus $H\cap K$ is nonempty. Now, since $H\leq G$ and $K\leq G$ we know that $H$ and $K$ are both subsets of $G$. Therefore, by the definition of the intersection of two subsets it is proven that $H\cap K$ is a nonempty subset of $G$.

Now it must be shown that $H\cap K$ is closed under the binary operation of G. Let $a,b\in H\cap K$. Thus, $a,b\in H$ and $a,b\in K$. Since $H\leq G$ we know that $ab\in H$. Similarly, $ab\in K$. Thus, by the definition of intersection $ab\in H\cap K$. This proves that $H\cap K$ is closed under the binary operation of $G$.

We must now show that $H\cap K$ is closed under taking inverses. Let $a\in H\cap K$. It has already been proven that if $H\leq G$, then $a^{-1}\in H$. Similarly, $a^{-1}\in K$. Thus, by the definition of intersection $a^{-1}\in H\cap K$. This mean $H\cap K$ is closed under taking inverses.

Hence, by the subgroup test $H\cap K\leq G$.

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