Show every left coset is equivalent to the right coset Assume that H is a subgroup of index 2 of finite group G.  I'm asked to show that every left coset is a right coset of H.  
I first see that the number of left cosets of H in G is 2.  I guess I am then told that H must have half as many elements as G then it shows that left equals right coset.  
this was the answer I was given but for a first time abstract algebra student this is incredibly hard to see.  Can someone help me visualize this, possibly with an example?
 A: The key is in understanding a fact about cosets that are true for both left and right cosets:
The set of cosets of $H$ in $G$ partition $G$ (i.e., every element of $G$ belongs to some coset of $H$.
With this in mind, our conclusion is simple. Consider the set of left cosets of $H$ in $G$. We know that $1H = H$ is one particular left coset. Now consider $G-H$, the set of elements in $G$ which are not in $H$. By our fact, the elements of $G-H$ belong to some coset of $H$ in $G$. There are only two cosets, since the index of $H$ in $G$ is two. Since they are not in $H$, the elements of $G-H$ must belong to the second left coset of $H$ in $G$. Hence, the two left cosets of $H$ in $G$ are therefore $H$ and $G-H$. 
Similarly, we can observe that $H1 = H$ is a right coset in $G$. By the same exact reasoning as above, the only other possible right coset of $H$ in $G$ is $G-H$. The two right cosets of $H$ in $G$ are therefore $H$ and $G-H$. The left and right cosets hence coincide. 
Note that this is really a combinatorial principle of sorts at work. If I tell you that I have a basket full of apples and oranges, and I pull out a piece of fruit and tell you that it is NOT an apple, it must be an orange! 
A: There are only two cosets of $H$. Let $g\not\in H$. Then, we can write the two left cosets as $gH$ and $H$. The right cosets, similarly, are $H$ and $Hg$. 
Clearly $H=H$, so what do you suppose is the relation between $gH$ and $Hg$?
A: The elements of $H$ form a coset corresponding to $1$ whichever side you choose. $H$ contains half the elements in the group (index $2$). All cosets are the same size. The other coset must contain precisely the elements not in $H$ whichever side you choose.
