I need to prove the following fact: if $G$ is a non abelian group of order $6$, then $G$ is isomorphic to the dihedral group $D_3$. Here what I have done:
Suppose that $G$ is non abelian group of order $6$. Since $G$ is not abelian, it cannot have a element of order $6$. (This is correct, since if $G$ had a element of order $6$, then $G$ would be cyclic, and hence abelian, a contradiction). Also, it cannot have elements of order 1, by the same argument as before.
But I got stucked in here. I'm struggling with the two other possibilities of the order of the elements in $G$: order $2$ and $3$. Any help or hints that you can bring me would be appreciated.
Best Answer
If you're allowed to use Cauchy's theorem, you may argue as follows:
Since $2$ and $3$ are prime numbers that divide $6$, we know that there are elements $a$ and $b$ of orders $2$ and $3$ respectively. Therefore, $\{e,a,b,b^2,ab,ab^2\} \subseteq G$. Since $G$ has $6$ elements, this subset is all of $G$.
Now the only thing left to do is to see how $a$ and $b$ interact with each other. Remember that $a=a^{-1}$ and $b^2=b^{-1}$. Now consider $aba$:
$aba=e \implies b=e$. Contradiction
$aba=a \implies b=a$. Contradiction
$aba=b \implies ab=ba$. But $G$ is assumed not to be Abelian. Contradiction.
$aba=ab \implies a=e$. Contradiction
$aba=ab^2 \implies a=b$. Contradiction
Hence, $aba=b^2$. This means that $aba=b^{-1}$. Hence, $G$ is isomorphic to $D_3$.