Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 7 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.