There are only two types of groups of order $6.$
Could anyone advise on how to prove a/m claim? Here is my attempt but I'm stuck:
If $\exists g\in G$ such that $o(g) =6,$ then $G = \left \langle {g}\right \rangle.$
If not, let $G = \{g_1,g_2,g_3,g_4,g_5,e\},$ where $o(g) \in \{2,3\} , \forall g\in G-\{e\}$
Also, $\exists i \in \{1,2,3,4,5\}$ such that $o(g_i)=2.$
Best Answer
There's an element $a$ of order two and an element $b$ of order three (Cauchy). If they commute, then $ab$ is of order $6$ and $G$ s cyclic. Otherwise, the elements $1,a,b,b^2,ab,ba$ are pairwise distinct. One of them must be $ab^2$ and $ba$ is the only candidate for that. This determines $G$ completely.