Proof about the isomorphism of a non abelian group and the dihedral group

Question

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.

Answer 1

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 $D_3$.

Answer 2

So you know that there is a subgroup of order 2 in $G$, call it $H$. We can define a map $G \to S_3$ (or $D_3$ but here it's more useful to think of it as $S_3$) by looking at group actions on the different cosets of $H$. Anything that fixes all cosets must be in $H$. Now kernels of maps are always normal but I claim that $H$ is not normal. Can you prove this step? This implies that the kernel cannot be all of $H$, i.e it must be just the identity. Now we have an injective homomorphismm between two finite groups, so they must be isomorphic.