Return to Homework 5, Glossary, Theorems

### Problem 9

(a) Show that a nonempty subset $H$ of a group $G$ is a subgroup of $G$ if and only if $ab^{-1} \in H$ for all $a,b \in H$.

(b) Let $H$ be a subgroup of a group $G$. For $a,b \in G$, define $a \sim b$ if and only if $ab^{-1} \in H$. Show that $\sim$ defines an equivalence relation.

**Solution**

(a)

We need to show that

if $H \leq G$ with $a,b \in H$, then $ab^{-1} \in H$

and that

if for all $a,b \in H, ab^{-1} \in H$ as well, then $H \leq H$.

Suppose that $H \leq G$ with $a,b \in H$.

Then $b^{-1} \in H$, by definition of subgroup.

Hence, $ab^{-1} \in H$, since subgroups are closed under the operation.

Now, let $a,b \in H$ and suppose $ab^{-1} \in H$.

(b)

**Reflexive**

Suppose $a \in H$. Now, $aa^{-1}$ is in $H$ because $aa^{-1}=e$ and since $H$ is a subgroup, $H$ must contain the identity. Hence, $a \sim a$.

**Symmetric**

Suppose $a \sim b]]. Then [[$ab^{-1} \in H$. Since $H$ is a subgroup, $(ab^{-1})^{-1} \in H$. That is, $a^{-1}b \in H$. Hence, $b \sim a$.

**Transitive**

Suppose $a\sim b$ and $b\sim a$.

Then $ab^{-1} \in H$ and $bc^{-1} \in H$.

Since $H$ is closed by definition of subgroup $(ab^{-1})(bc^{-1}) \in H$.

By associativity, $a(b^{-1}b)c^{-1} \in H$.

That is, $ac^{-1} \in H$.

Hence, $a \sim c$.