Prove that, A non- empty subset $\mathrm{H}$ of a group $(\mathrm{G}, *)$ to be a subgroup is $\mathrm{a} \in \mathrm{H}, \mathrm{b} \in \mathrm{H} \rightarrow \mathrm{a}^* \mathrm{~b}^{-1} \in \mathrm{H}$

Solution:

Let H be a subgroup of the group G and $b \in H, a \in H$ then $b \in H$ ie. $b^{-1} \in H \cdot$ [Existence of inverse]

$\therefore \quad a \in H, b \in H, \quad b^{-1} \in H, a^{-1} \in H$

$\Rightarrow a b^{-1} \in H[B y$ closure property]

Coversly_:-

suppose the given condition is true in H.

$ \therefore \quad H=\phi $

Let $a \in H$.

Therefore, Identity element exists.

Again, by the same condition -

$ \begin{aligned}\\ & e \in H, \quad a^{-1} \in H \\\\\ \Rightarrow & e a^{-1}=a^{-1}\\ \end{aligned}\\ $

ie. H is the element for every inverse element in H.

Finally, $a \in H, \quad b \in H$

ie. $a \in H, b^{-1} \in H$

So, H is closed for the operation of G.

Hence, H is a subgroup of G.