# A non-abelian group of order $6$ is isomorphic to $S_3$

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 $S_3$.

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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

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 $S_3$.

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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

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 $S_3$.