$[G:H] < \infty$ then $gHg^{-1} \subset H$ and is it true that $gHg^{-1} = H$ G is a group. $H < G$  and $ g \in G$
$gHg^{-1} \subset H$
I need to prove the following :
a) if $[G:H] < \infty$  then $gHg^{-1} = H$
b) without the additional fact given in (a) is it true that $gHg^{-1} = H$
What I tried : 
a) from $gHg^{-1} \subset H$ 
i know that $gH \subset Hg$
and i can make one to one map between gH  - > Hg 
so I know they have the same size (less then infinity) 
so I know they are equal ? is that correct? 
b) here I'm having harder time - I assume it is not correct and I tried thinking about 
$H = g^{-1}M$  (where M is some other group) 
then I get $M \subset g^{-1}Mg$ 
which can make sense that is not correct the other way if $M < G$.
but I think I'm wrong here.. 
Any help on both will be very appriciated
 A: Hint: you know that for all $a\in G$ you have $$aHa^{-1}\subset H$$ so as you correctly point out, this gives $$aH \subset Ha. $$  Now look at the first expression for $a=g$ and $a=g^{-1}$.  What inclusions do you get?
Edit: In case you are only assuming $gHg^{-1}\subset H$ for one $g\in G$, here is an extra hint on how to proceed if $[G:H]<\infty$.
1) Show that as $a$ ranges over $G$, the number of distinct groups $aHa^{-1}$ is finite.
2) Consider the chain of groups $H_n = g^nHg^{-n}$.  Show that $H_{n+1}<H_n$ for all $n\geq 0$.
3) Consider two cases: either the chain stabilizes, i.e. $H_{n+1}=H_n$ for all $n$ bigger than some $n_0$ (if $H_m=H_n$ for some $m>n$ then also $H_{n+1}=H_n$ because the groups are nested - this also takes care of the case $g^k=e$ for some $k$), or there are infinitely many distinct $H_n$.  In the first case, from $H_{n+1}=H_n$ conclude that $H_1 = H$ and you are done.  In the second case arrive at a contradiction using 1).
For the case of infinite index, it is not generally true that $gHg^{-1}=H$.  As a counterexample, consider the following:
The cyclic subgroup $H$ of $\textrm{GL}(2,\mathbb{Q})$ generated by $\pmatrix{1 & 1 \\0 & 1} $ is isomorphic to the additive group $\mathbb{Z}$: $H = \{\pmatrix{1 & z \\0 & 1}|z\in \mathbb{Z}\}$.  Consider $g= \pmatrix{2 & 0 \\0 & \frac{1}{2}} $.  Then $gHg^{-1} = \{\pmatrix{1 & 4z \\0 & 1}|z\in \mathbb{Z}\}\subsetneq H$.
To answer pid's comment: We know that there are finitely many left cosets $aH$ and finitely many right cosets $Ha$ (and their number is the same).  Pick $a_1,\cdots, a_m$ and $b_1,\cdots, b_m$ to be representatives of left and right cosets respectively (you can pick as right representatives $b_j=a_j^{-1}$ but I will not use this information here).  So every left coset is of the form $a_iH$ for some $i\leq m$ and every right coset of the form $Hb_j$ for some $j\leq m$.  
Also observe that since $H$ is a subgroup, $H\cdot H=H$, and recall associativity. Now suppose the coset $gH$ coincides with the coset $a_iH$ and the coset $Hg^{-1}$ coincides with the coset $Hb_j$.  Then $$gHg^{-1} = gH\cdot Hg^{-1} = a_iH\cdot Hb_j. $$
So we see that all these groups $gHg^{-1}$ are product sets of specific pairs of left coset/right coset, and those are finite in number, at most $m^2$.
