# There are only two types of groups of order $6.$

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 $o(g_i)=2.$

• The notation $o(g)=2\vee3$ is very unusual (and, strictly speaking, wrong). $2$ and $3$ aren't propositions, they're numbers. I thought that was a GCD or something! Oct 27, 2013 at 16:32
• Duplicate...? Oct 27, 2013 at 16:35
• And here as well. Or there. Oct 27, 2013 at 16:37
• @JackM: $o(g)=2\lor3$, using mathematical symbols to represent the linguistic idiom "the order of $g$ is $2$ or $3$" is a confusing abbreviation. I have changed it to $o(g)\in\{2,3\}$.
– robjohn
Oct 27, 2013 at 17:21

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.
• Thank you! Is it possible to show that there's an element $a$ of order two and an element $b$ of order three without using Cauchy theorem? Oct 27, 2013 at 17:34
• Hello Hagen. I do not understand how this completely determines $G$. Can you please explain? Thank you. Jun 28, 2017 at 17:07