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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 $o(g_i)=2.$

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    $\begingroup$ 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! $\endgroup$
    – Jack M
    Oct 27, 2013 at 16:32
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    $\begingroup$ Duplicate...? $\endgroup$
    – Julien
    Oct 27, 2013 at 16:35
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    $\begingroup$ And here as well. Or there. $\endgroup$
    – Julien
    Oct 27, 2013 at 16:37
  • $\begingroup$ @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 $o(g)\in\{2,3\}$. $\endgroup$
    – robjohn ♦
    Oct 27, 2013 at 17:21

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Alexy Vincenzo
    Oct 27, 2013 at 17:34
  • $\begingroup$ If the group is not cyclic, then by Lagrange the elements must be of order 2or 3. Just show they can't all be order 2. $\endgroup$
    – Bill Kleinhans
    Oct 27, 2013 at 21:00
  • $\begingroup$ Hello Hagen. I do not understand how this completely determines $G$. Can you please explain? Thank you. $\endgroup$
    – Cauchy
    Jun 28, 2017 at 17:07

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