$\{1,-1,i,i^2\}=G$ is a finite cyclic group and $H =\{1,-1\}$ is a Subgroup of G.
Order of $G$ is $4$
And order of $H$ is $2$
$G\cap H =\{1,-1\}$
$P(G \cap H)= ((\text{number of elements of } G \cap H)/\text{number of elements of }G)=2/4=1/2$
Here $G$ is sample space and $H$ is sub set of $G$ .So $H$ is an event of $G$.
$P(G|H)=P(G \cap H)/P(H)=(1/2)/(1/2)=1
Is it wrong ?
If $H\subseteq G$ or equivalently $G\cap H=H$ then if $P(H)>0$: $$P(G\mid H)=\frac{P(G\cap H)}{P(H)}=\frac{P(H)}{P(H)}=1$$