I know that the group $G=D_3 \times D_3$; with $D_3$ is the Dihedral group of order 6; admits exactly $3$ subgroups of index $2$.
We can prove this by using the fact that $G$ and $G/G^2$ have the same number of subgroups Of index $2$. In our case we have $G/G^2 \simeq (\mathbb{Z}/2\mathbb{Z})^2$, so $G/G^2$ is a $\mathbb{Z}_2$-vector space of dimension $2$. The subgroups of index $2$ of $G/G^2$ are exactly the hyperplanes.
By group theory we have this known result:
"Every $n$-dimensional vector space over $\mathbb{Z}_p$ has exactly $(p^n-1)/(p-1)$ hyperplanes."
For our case we find $2^2-1=3$ hyperplanes of $G/G^2$.
Anyway, my question will be simpler than the data I mentioned above!
I know that The two subgroups of $G$ of index $2$ are $D_3 \times \mathbb{Z}/3\mathbb{Z}$ and $ \mathbb{Z}/3\mathbb{Z}\times D_3$ but I can't find the third one, even thought I know that its not going to be a subproduct of $D_3 \times D_3$.
So I'm looking for away to define this subgroup.
P.S. I thought about Goursat lemma but I find it very complicated.
Best Answer
If $P<D_3$ and $|P|=3$, then $D_3=P\cup Px$ for some $x\notin P$. We have $H_1=(P,D_3)$, $H_2=(D_3,P)$ and $H_3=(P,P)\cup (Px,Px)$ are all subgroups of index $2$.