Classify groups of order 6 by analyzing the following three cases: (i) G contains an element of order 6. 
(ii) G contains an element of order 3 but none of order 6. 
(iii) All elements of G have order 1 or 2.
I've got: 
(i) $G$ is the cyclic group of order 6, isomorphic to $\mathbb{Z}_6^+$. 
(ii) $G$ is $S_3$, the group of permutations of 3 objects. 
(iii) $G = \{1, x, y, z, xy, zx \}$ (what named group is this?)
 A: Up to isomorphism, every group of order $6$ is isomorphic to either $\mathbb Z_6$ or $S_3$. This can be proved directly without too much difficulty just by brute force constructions of the group multiplication table on six elements (though it's a bit lengthy). A quick way to see it is as follows. If $G$ has an element of order $6$ then it must be cyclic, thus isomorphic to $\mathbb Z_6$. So, assume no element of order $6$ exists. The order of each element must divide $6$, thus it's either $1,2$, or $3$. There is (always) precisely one element of order $1$, the identity. If there is one element of order $2$ and one of order $3$, then all elements are accounted for and finishing the multiplication table is now easy, giving a group isomorphic to $S_3$. By counting, having all non-identity elements have order $3$ will not add up to $6$ elements. Having all non-identity elements have order $2$, means the group is abelian and then attempting to complete the multiplication table will lead to a dead-end. 
A: You are almost correct, but the third case is impossible.  There can only be one element of order $1$ (the identity), so that leaves us $5$ elements of order $2$.
Such a group must be abelian ($xy = (xy)^{-1} = y^{-1}x^{-1} = yx$), and therefore generated by more than $2$ elements (otherwise, you have the Klein four group).
But if it's generated by distinct $x$, $y$, and $z$, then $xy$, $xz$, $yz$, and $xyz$ are all distinct as well (otherwise, two of $1,x,y,z$ would be equal by the cancellation law).  That gives us at least $8$ elements, which is a contradiction.
