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 know that it is duplicated. But I'm confusing some step of this proof. Please help me.

pf) Let $ G $ be a nontrivial group of order $ 6 $.

Since $ G $ is non-abelian, no elements in $ G $ have the order $ 6 $.

Assume that every element except $ e $ is of order 2.

If $ x $ and $ y $ are of order $ 2 $ and not equal. Then $ \langle x, y \rangle $ has the order $ 4 $. It is contradiction, since $ 4 $ does not divide the order of $ G $.

So, $ G $ must contain an element of order $ 3 $, say $ y $. Let $ \langle y \rangle$ and $x \langle y \rangle$ be two cosets.

Consider $ yx $.

Since $ x\notin\langle y \rangle$ and $ y\neq x$, $yx = xy$ or $yx=xy^2$ .

In the case of $yx = xy$, consider the order of $ xy$.

If the order of $ xy$ is $2$, then $y=x^2$. And so the order of $ x $ is $ 6 $ . Then it tis contradiction.

(I don't understand why it consider only when the order of $ xy $ is $ 2$ .)

Hence $yx=xy^2$. Moreover, since $ x^2 \in x \langle y \rangle$ , $ x^2 \in \langle y \rangle$ .

(This is another confusing part. Why $ x^2$ need to be in $ x \langle y \rangle$? And even if it is true, why it means $ x^2 \in \langle y \rangle$? )

Since $ x \neq y, y^2 $, $ x = e $. Hence $ G $ is isomorphic to $ S_3 $.

share|cite|improve this question
How do you conclude that it will have a subgroup of order $4$ is all the elements have order $2$? – Tobias Kildetoft Apr 25 '13 at 13:33
@TobiasKildetoft sorry, I skipped some step. If $ x $ and $ y $ are of order $ 2 $ and not equal. then, $ \langle x, y \rangle $ has the order 4. I editted the article. Thank you. – user73309 Apr 25 '13 at 13:39
It is not correct that the subgroup generated to two elements of order $2$ has order $4$ (indeed, $S_3$ has two distinct elements of order $2$, and they generate the entire group). To show not all elements have order $2$, I recommend showing that if they do, then the group is abelian. – Tobias Kildetoft Apr 25 '13 at 13:47
@TobiasKildetoft But it is true under the assumption that all elements (specifically $x, y$ and $xy$) have order $2$. – Arthur Apr 25 '13 at 13:47
@Arthur Indeed it is, but that last part is important for the argument. – Tobias Kildetoft Apr 25 '13 at 13:51
up vote 4 down vote accepted

Assume (for contradiction) that $xy = yx$.

When you consider the order of $xy$, it can only be equal to $2$ or $3$, because you've already argued that the group has no elements of order $6$ and $xy = e \Rightarrow x \in \langle y \rangle$, a contradiction.

If $xy$ had order $2$, then $e = (xy)^2 = x^2 y^2$ implies that $y = y^{-2} = x^2$ (using that $y$ has order $3$), forcing $x$ to have order $6$, a contradiction. To spell this out, clearly $x^6 = y^3 = e$, while if $x^k = e$, then $k \neq 2, 4$ because $y, y^2 \neq e$ while $k$ cannot be odd because in that case $x^k \in x \langle y \rangle$ which does not contain $e$.

Now suppose that the order of $xy$ is $3$. Then $e = (xy)^3 = x^3 y^3 = x^3$, so $x$ has order $3$. But $x^3 \in x \langle y \rangle$, a contradiction by the same argument as above.

This proves that $xy = yx^2$. If $x^2 \in x \langle y \rangle$, then $x \in \langle y \rangle$, which is false by hypothesis. So $x^2 \in \langle y \rangle$. We wish to show that $x^2 = e$, so we must rule out $x^2 = y$ and $x^2 = y^2$. If $y = x^2$, then as above, we argue that $x$ has order $6$, a contradiction. If $x^2 = y^2$, then $x^2$ has order $3$, so $x = x^4 = y^4 = y$, also a contradiction.

Once you know that $x^2 = y^3 = e$ and $xy = yx^2$, it is possible to construct an explicit isomorphism from $G$ to $S_3$.

share|cite|improve this answer
Michael, can you please clear some issues I have with this problem? The group has no elements of order $6$ because then that element will be cyclic but cyclic groups are abelian which is a contradiction, correct? Also, how did you get that $y = y^{-2}$? And how will that force $x$ to have order $6$? – Lays Oct 19 '13 at 8:23
I got $y = y^{-2}$ from your choice of $y$ as an element of order $3$. (Multiply the relation $y^3 = e$ by $y^{-2}$ on both sides.) To get that this forces $x$ to have order $6$, use the fact that $x$, $x^2 = y$, $x^3 = xy$ are all not the identity, so $x$ does not have order $1$, $2$, or $3$. Therefore, since the order of $x$ divides 6, it must be equal to 6. – Michael Joyce Oct 19 '13 at 17:14
Thank you very much! – Lays Oct 20 '13 at 3:04

Here is an alternative answer:

By Cauchy's theorem, a group of order 6 has an element $x$ of order $2$ and an element $y$ of order $3$. These two elements generate the group. The 6 elements $e$, $y$, $y^2$, $x$,$xy$, $xy^2$ must all be different from each other, hence this is the list of all elements of the group. Therefore, $yx$ must be somewhere on this list.

Checking each element: we know that $yx\neq e$ because $x\neq y^{-1}$, $yx\neq y$ because $x\neq e$, $yx\neq y^2$ because $x\neq y$, $yx\neq x$ because $y\neq e$, and $yx\neq xy$ because by assumption the group is not abelian. Thus $yx=xy^2$, hence our group is the symmetric group $S_3$.

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.