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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.


marked as duplicate by Dietrich Burde abstract-algebra Jan 2 '18 at 10:13

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  • $\begingroup$ "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. $\endgroup$ – Arthur Jan 2 '18 at 8:12
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    $\begingroup$ If all elements had order 2 then the group would be abelian: $1 = (ab)^2 = abab$ so $bab = a^{-1} = a$ and hence $ba = ab^{-1} = ab$. $\endgroup$ – Tob Ernack Jan 2 '18 at 8:12
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    $\begingroup$ 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. $\endgroup$ – S.C.B. Jan 2 '18 at 8:14
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    $\begingroup$ 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. $\endgroup$ – Tob Ernack Jan 2 '18 at 8:19
  • $\begingroup$ @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 $a^p=e$. $\endgroup$ – ZFR Jan 2 '18 at 8:55

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 $S_3$.

  • $\begingroup$ Let me ask you the first question: Why $a$ is not power of $b$ and $b$ is not a power of $a$? $\endgroup$ – ZFR Jan 2 '18 at 10:12
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    $\begingroup$ Otherwise they would commute!!! And $ab \neq ba$. $\endgroup$ – Nicky Hekster Jan 2 '18 at 10:13
  • $\begingroup$ Indeed, very elegant and short solution! Nice answer! +1. I accept it as the best answer :) $\endgroup$ – ZFR Jan 2 '18 at 10:47
  • $\begingroup$ Thanks!! Yes you do not have to use Cauchy, Lagrange, Sylow or whoever ... $\endgroup$ – Nicky Hekster Jan 2 '18 at 11:06

If you do not allow the use of Sylow theorem, Cauchy's theorem or group actions, then you must construct by hand the multipilcation 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?


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