Question

Prove that a nonempty subset Hof a group G is a subgroup of G if and only if whenever $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{h}$, then$\mathit{a}{\mathit{b}}^{\mathbf{-}\mathbf{1}}\mathbf{\in }\mathit{H}$ .

Short answer

Itis proved that a non-empty subset H of a group G is a subgroup of G.

Step-by-step solution

Subgroup Theorem

A non-empty subset $H$of a group $G$is a subgroup of the latter if the elements $a$and $b$belong to $H$, then the element belongs $ab$to $H$, and if the $a$element belongs to$H$ , then $a^{-1}$the element belongs to$H$ .

Show that a non-empty subset H of a group G is a subgroup of G

First, consider that set$H$is a subgroup of $G$. This implies that for two arbitrary elements $a$and$b$in the subgroup $H$, the product $ab$is also present in $H$, and the inverse of every element is also present in group $H$.

Therefore, the element $a{b}^{-1}$must be present in group $H$.

Now, consider for two arbitrary elements $a$and$b$in set$H$, the product$a{b}^{-1}$is also present in $H$. Set $H$is non-empty.

Suppose$H$ has an element $h$.

Take the values $a=h$and$b=h$ , which will give $h{h}^{-1}$and in turn, the identity element $e$. Therefore, this implies that the set$H$has an identity element.

Take the values$a=e$ and $b=h$to prove the existence of inverse. This will give$e{h}^{-1}$ and in turn, the inverse element $h^{-1}$. This implies that set $H$has an inverse element for every element present in it.

Take two elements $a$and $b$in $H$and another element $c=b^{-1}$in$H$ to prove that set$H$ satisfies the closure property.

Find $a{c}^{-1}$as follows:

$$\begin{gathered} a{c}^{-1}=a{\left(b^{-1}\right)}^{-1} \\ =ab \\ \in H \end{gathered}$$

It is observed that the set $H$satisfies the closure property.

From the above results, it can be said that set $H$is a subgroup of$G$ .

Therefore, it is proved that a non-empty subset H of a group G is a subgroup of G.