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If $G$ is a non-abelian group of order $6$, prove that $G\cong S_3$.

I have met this problem in forum but it's solution is somewhat brief and not detailed and I cannot understand some its moments.

My efforts: Looking at the structure of $S_3$ I tried to draw up similarities between non-ableian group $G$ and symmetric group $S_3$.

Since $G$ has order $6$ then none of the elements have order $6$, otherwise it would be cyclic then abelian. Hence, all elements of $G$ except $e$ have order $2$ or $3$.

The case when all elements have order $2$ is not possible. I do not know why it is true. Please explain that.

Then one of the element has order $3$. And what to do after that I do not know.

It would be great and very useful if somebody will demonstrate the whole and detailed proof.

Remark: This problem from Herstein's book and please do not use Sylow and Cauchy's theorem and group actions.

RFZ
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  • "The case when all elements have order $2$ is not possible. I do not know why it is true. Please explain that." Caychy's theorem says that since $3$ is a prime which divides $6$, the group must have at least one element of order $3$. That and Sylow's first theorem are as close we can get to a converse of Lagrange's theorem. – Arthur Jan 02 '18 at 08:12
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    If all elements had order 2 then the group would be abelian: $1 = (ab)^2 = abab$ so $bab = a^{-1} = a$ and hence $ba = ab^{-1} = ab$. – Tob Ernack Jan 02 '18 at 08:12
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    Suppose all elements have order $2$. Then we would have that $\phi(a)=a^{-1}$ is the identity map and thus an isomorphism from $G$ to itself. This would imply that $$ab=\phi(a)\phi(b)=\phi(ab)=b^{-1}a^{-1}=ba$$ And thus abelian. – S.C.B. Jan 02 '18 at 08:14
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    You can also conclude that there is an element of order 2 because otherwise you could pair the elements as $(a, a^{-1})$ except for 1, and so the order would have to be odd. – Tob Ernack Jan 02 '18 at 08:19
  • @Arthur, however the Cauchy's theorem in my book refers to finite abelian groups. "Suppose $G$ is a finite abelian group and $p\mid o(G)$, where $p$ is a prime number. Then there is an element $a\neq e$ such that $a^p=e$. – RFZ Jan 02 '18 at 08:55
  • @RFZ That is a needlessly weak version. But sure, if Cauchy doesn't work, the other two comments give excellent reasons. – Arthur Jan 02 '18 at 08:57
  • @TobErnack, Thanks a lot! I have understood that case but how to continue if there is one element of order $3$? – RFZ Jan 02 '18 at 09:03
  • @DerekHolt, Thanks I have done that. Just now I have difficulties with the second case when one of elements has order $3$. – RFZ Jan 02 '18 at 09:11
  • OK, if $a$ has order $3$, then $\langle a \rangle$ is a subgroup of index $2$, so it is normal. Let $b \in G \setminus {a}$. Then $b^2 \in \langle a \rangle$. If $b^2 = a$ or $a^2$ then $G$ is cyclic. If $b^2=1$ then $b^{-1}ab = a$ or $ a^{-1}$ giving the cyclic or dihedral group. You should complete the details yourself. – Derek Holt Jan 02 '18 at 09:13
  • @DerekHolt, Here are my own efforts. Suppose $y$ is an element of order $3$ then $\langle y \rangle$ cyclic subgroup of order 3. Then $\langle y \rangle$ and $x\langle y \rangle$ are two right cosets and their union is desired group $G$. Thus $G={e,y,y^2,x,xy,xy^2}$ but we know that $S_3={e,\phi, \psi,\psi^2, \phi\psi, \psi\phi}$. If I'll show that $xy^2=yx$ in $G$ then the rest is easy. I have problems with showing that $xy^2=yx$. – RFZ Jan 02 '18 at 09:18
  • The solution of Mark Bennet at the duplicate is completely elementary. – Dietrich Burde Jan 02 '18 at 10:14
  • @DerekHolt, We know that $G=\langle y\rangle \sqcup x\langle y\rangle$ where $o(y)=3$. Also $x\in x\langle y\rangle$. Consider the element $yx$, then $yx\notin \langle y\rangle$ hence $yx\in x\langle y\rangle$ we have two cases: $yx=xy$ and $yx=xy^2$. Let's consider the first case. If $xy=yx$ and since $o(xy)\mid 6$ then $o(xy)\in {1,2,3,6}$. If $o(xy)=1$ then $xy=e$ so $x\in \langle y\rangle$ which is impossible. If $o(xy)=2$ then $e=(xy)^2=x^2y^2$ so $x^2=y^{-2}=y\neq e$ and $x^3=xy\neq e$. So $o(x)=6$ which is not possible. – RFZ Jan 02 '18 at 12:00
  • @DerekHolt, If $o(xy)=3$ then $e=(xy)^3=x^3y^3=x^3$ since $y^3=e$. Thus $x^3=e\in \langle y\rangle$. How to show that also $x^3\in x\langle y\rangle$ in order to derive contradiction? – RFZ Jan 02 '18 at 12:00
  • This question got duped up a chain of older questions, all of which use results that are not wanted here. That was not a good idea... – Mariano Suárez-Álvarez Jan 29 '18 at 19:58

2 Answers2

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This can be dealt with by completely elementary means. Assume that $G$ is non-abelian and has order $6$. Hence we can find two elements $a,b \in G$ with $ab \neq ba$ and $1 \notin \{a,b\}$ (here $1$ denotes the identity element of $G$ and note that $a$ is not a power of $b$ and $b$ is not a power of $a$). So the subset $\{1,a,b,ab,ba\}$ of $G$ consists of exactly five different elements.

Let us have a look at $a^2$. Then an easy check gives $a^2 \notin \{a,b,ab,ba\}$. Hence $a^2=1$ or $a^2$ is a "new" element $\neq 1$.

Assume that $a^2 \neq 1$, so $G=\{1,a,b,ab,ba,a^2\}$. Then $a^3 \notin \{a,b,ab,ba,a^2\}$. So in this case we must have $a^3=1$. And also $b^2 \notin \{a,b,ab,ba,a^2\}$ (for the last element in this set: if $b^2=a^2$ then $ab^2=a^3=1$, hence $a^{-1}$ and thus $a$ is a power of $b$, contradiction), so $b^2=1$. Now the map $a \mapsto (1 2 3)$ and $b \mapsto (1 2)$ yields an isomorphism with $S_3$.

Because of symmetry, we can finally assume that $a^2=1=b^2$, that is $a=a^{-1}$ and $b=b^{-1}$. It is again easily checked that $aba \notin \{1,a,b,ab,ba\}$. Then $aba$ is the sixth element and by symmetry, $aba=bab$, so $ba=(ab)^2$ and $(ab)^3=1$. Now the map $ab \mapsto (1 2 3)$ $b \mapsto (12)$ gives the desired isomorphism with $S_3$.

Nicky Hekster
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If you do not allow the use of Sylow theorem, Cauchy's theorem or group actions, then you must construct by hand the multiplication table of a group of order $6$, assuming it is not abelian (which rules out the cyclic case). Then, you must compare your multiplication table to that of $S_3$ and see that they are the same.

The question is: do you really want to do that?

J. W. Tanner
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uniquesolution
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