Show that if $G$ is a finite group and $H_i$ are subgroups of $G$ with $[G:H_i]=2$ then $[G:\cap H_i]=$ some power of $2$.
Let the number of subgroups of $G$ be $H_1,H_2,\ldots ,H_m$
Its clear that each $H_i$ is a normal subgroup of $G$ and every $H_i$ has exactly two left/right cosets.
Let the left cosets of $H_1$ in $G$ be $H_1,g_1H_1$ ,that of $H_i$ in $G$ be $H_i,g_iH_i$ and so on.
Let $H=\cap H_i$
Now we know that $[G:H\cap K]\le [G:H][G:K]$ for any two subgroups $H,K$ of $G$.
Thus we have
Now I need to show only that $[G:H]\ge 2^m$
Now I understand that since $g_i\notin H_i\implies g_i\notin H$
hence we have at least $m$ cosets of $H$ in $G$ given by $g_1H,g_2H,\ldots g_mH$
But I need to find at least $2^m$
How can I do it?
Please give some hints