Coset representation ; $\{ Hg_i \} = \{ Hg_i^{-1} \}$ This is my simple curiosity 
If $H$ is normal in $G$ then $H\setminus G$ is a group : $$ Ha \cdot Hb = H ab $$
  But if $H$ is not normal then the coset space $H\setminus G$ is not a group, but we guess that there exists a some property which can be satisfied in a group.
Let $H$ be a subgroup of a finite group $G$. Then we have $H\setminus G = \{ Hg_i \}_{i=1}^m$ where $g_i$ is representative and $m=[G:H]$. Here 

Question : $$ \{ Hg_i \}_{i=1}^m = \{ Hg_i^{-1} \}_{i=1}^m$$ 

That is, $Hg_i=Hg_j^{-1}$ for some $j$. This holds ?
If not, $Hg_i^{-1}=Hg_j^{-1}$ for some $i\neq j$ so that $g_i^{-1} g_j\in H$. I cannot prove or disprove.  
Specific Case : $$G=N\times_\phi H,\ N=\langle a\rangle={\bf Z}_p,\ H=\langle b \rangle={\bf Z}_q\subset {\rm Aut}\ (N)\ (q|(p-1))$$
where $p$ and $q$ are primes. 
Then there exists $x\in G$ s.t. $H'=\langle x\rangle $ has order $q$ and $ H'\setminus G = \{ H'c^k \}_{k=0}^{p-1}$ where $c\in G$ has order $p$. In this case the above equality holds.  
 A: Not every right transversal to $H$ in $G$ has that property, but it is always possible to choose a particular right transversal with this property. There is a Theorem of P. Hall which asserts that whenever $H$ is a subgroup of a finite group $G,$ there is a right transversal $T$ to $H$ in $G$ which is also a left transversal. 
That is, if $T = \{t_{i}: 1 \leq i \leq n \},$ where $n = [G:H],$ then we have $G = \bigcup_{i=1}^{n} Ht_{i}$  and $G = \bigcup_{i=1}^{n}t_{i}H$. Then for $i \neq j,$ we have $t_{i}t_{j}^{-1} \not \in H$ and $t_{i}^{-1}t_{j} \not \in H.$ The second condition tells us that the right cosets $Ht_{i}^{-1}$ and $Ht_{j}^{-1}$ are distinct. Hence we do have the equality asked for in the question for this particular transversal.
Here is an explicit counterexample to the necessary disjointness for a general transversal. Take $G = S_{4}$ and $H$ to be the stabilizer of $4$ in $G,$ which is the natural copy of $S_{3}$ inside $G.$ Note that the cosets $H(134)$ and $H(234)$ are distinct, because $(134)(432) = (412) \not \in H.$ However, the cosets $H(431)$ and $H(432)$ are equal, because $(431)(234) = (123) \in H.$  
A: If $G=S_3$ and $H=\{1,(12)\}$ take $g=(123)$ so $g^{-1}=(132).$
Then $Hg=\{(123),(13)\}$ while $Hg^{-1}=\{(132),(23)\}.$
[note multiplied left-to right here.]
Added: If the number of cosets is a finite $m$, then your list of cosets $Hg_i$ for $1 \le i \le m$ is the same as the number of cosets $Hg_i^{-1}$. The latter are also cosets, and are pairwise disjoint by the fact that the $Hg_i$ are pairwise disjoint. So both lists of cosets are complete lists of the cosets of $H$ in $G$. For example consider some fixed coset $Hg_j^{-1}.$ Then $g_j^{-1}$ must lie in one of the cosets $Hg_i$, and for that $i$ we have $Hg_i=Hg_j^{-1}.$
Note the following does not show disjointness of the cosets with representatives taken as the inverses. The question remains open.
A comment asked for why the $Hg_i^{-1}$ are disjoint. Suppose $Hg_1^{-1}=Hg_2^{-1}.$ Multiplying by $g_1$ on the right then gives $H=Hg_2^{-1}g_1,$ so that $g_2^{-1}g_1\in H$. But then $g_1,g_2$ are in the same coset of $H$, i.e. $Hg_1=Hg_2,$ against the choice of the coset reps for the cosets $Hg_i.$
