Any nonabelian group of order $6$ is isomorphic to $S_3$?

I've read a proof at the end of this document that any nonabelian group of order $6$ is isomorphic to $S_3$, but it feels clunky to me.

I want to try the following instead:

Let $G$ be a nonabelian group of order $6$. By Cauchy's theorem or the Sylow theorems, there is a element of order $2$, let it generate a subgroup $H$ of order $2$. Let $G$ act on the quotient set $G/H$ by conjugation. This induces a homomorphism $G\to S_3$. I want to show it's either injective or surjective to get the isomorphism.

I know $n_3\equiv 1\pmod{3}$ and $n_3\mid 2$, so $n_3=1$, so there is a unique, normal Sylow $3$-subgroup. Also, $n_2\equiv 1\pmod{2}$, and $n_2\mid 3$, so $n_2=1$ or $3$. However, if $n_2=1$, then I know $G$ would be a direct product of its Sylow subgroups, but then $G\cong C_2\times C_3\cong C_6$, a contradiction since $G$ is nonabelian. So $n_2=3$. Can this info be used to show the homomorphism is either injective or surjective? Thanks.

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You can't talk about the quotient $G/H$ unless you know that $H$ is normal; and you have no warrant for asserting that $H$ is normal. And if you try acting on the cosets by conjugation, you will find that it is not well-defined. – Arturo Magidin Jun 4 '12 at 4:22
This isn't your question, but the proof I expected to see in your document, and didn't see, is that (by Cauchy's theorem) $G$ has elements $a$ of order 2 and $b$ of order 3. The set $\{1, a, b, ab, b^2, ab^2\}$ is easily seen to exhaust $G$; no two of these can be equal if $a$ and $b$ are to have orders 2 and 3. Since $ba\in G$, we have $ba = a^ib^j$ for some $i\in\{0, 1\}$ and $j\in\{0,1,2\}$. $i=0$ and $j=0$ are easily ruled out. This leaves $ba=ab$ and $ba=ab^2$. The first is abelian. The second is the canonical presentation of $D_6 = S_3$, so we're done. – MJD Jun 4 '12 at 4:22
@TiffanyHwang: You are just acting on the wrong thing in the wrong way. You can use the cosets and left multiplication. – Arturo Magidin Jun 4 '12 at 4:27
@ArturoMagidin So if the action is left multiplication, then since necessarily $gH=H$ for $g$ in the kernel $K$, so $K\subset H$. But if $K=H$, $H$ is normal, then $n_2=1$, and I get the same problem as before. So $K=\{1\}$, and it's injective? Thanks. – Tiffany Hwang Jun 4 '12 at 4:30
@Tiffany: Indeed. – Arturo Magidin Jun 4 '12 at 4:31

You can't talk about the quotient $G/H$ unless you first prove that $H$ is normal (which you won't be able to do, since a group of order $6$ always has a normal $3$-subgroup, and if it has a normal $2$-subgroup then it is abelian). If you are trying to talk about the cosets of $H$ in $G$, then the action by conjugation is not well-defined, since the coset $H$ is not mapped to a coset of $H$ under conjugation by any element not in $H$ (precisely because $H$ is not normal).
If you want to use actions, you can do it: let $H$ be a subgroup of order $2$ and consider the action of $G$ on the left cosets of $H$ in $G$ by left multiplication. This gives you a homomorphism $G\to S_3$; the kernel is contained in $H$, but since $H$ is of order $2$ and not normal, that means that the kernel is trivial, and so the map is an embedding. Since both $G$ and $S_3$ have order $6$, it follows that the map is an isomorphism.