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In an attempt to prove that every group $G$ of order 6 is isomorphic to either $\mathbb{Z}_6$ or $S_3$, I stumbled upon one peculiar issue.

We can use Cauchy's Theorem to argue that since $|G|=3\cdot2$, $G$ must necessarily have elements of orders 1, 2, and 3. Another possibility is that there exists an element of order 6 in $G$. Now, suppose this is true, then $G$ is cyclic of order 6, which implies that $G\cong \mathbb{Z}_6$.

Alternatively, suppose that $G$ contains only elements of orders 1, 2, and 3. Then... (and here's where it appears to be the most difficult part)

There must necessarily be only two elements of order 3. For if there are more than two elements of order 3 then for some $a\in G$ s.t. $a^3=1$, $a^2=b$, where $|b|=2$ or $|b|=3$. Suppose that $b$ has order 2. Then $a^2a=1$ implies that $a^2 = a^{-1}$, thus $|b|\ne 2$. We conclude that $b$ has order 3. Then $a_1^2 = b_1$, $a_2^2=b_2$, $a_3^2=b_1$ or $a_3^2=b_2$. But this implies that $a_3=a_1$ or $a_3 = a_2$. Hence, $G$ must contain two elements of order 3.

We can now define a bijective homomorphism between $G$ and $S_3$, which implies that $G\cong S_3$.

I'm wondering, however, if there's a simpler way to prove that there must necessarily be exactly two elements of order 3 in $G$.

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    $\begingroup$ @MattSamuel I don't think the argument about "room" is compelling. You must prove that even if there exists two elements of order 3 ($a$ and $b$), that there cannot be two other elements of order 3 ($c$ and $d$). The statement may be true, but one needs a rigorous proof of it. Frankly, I think the proof in the question itself is simple enough, and I don't see a simpler argument. $\endgroup$ – David G. Stork Dec 11 '15 at 1:49
  • $\begingroup$ @Matt Samuel How do you see that? I think problematic observation is that if $g$ is order $3$ element then $g^2$ is also a order $3$ element. $\endgroup$ – user160738 Dec 11 '15 at 1:50
  • $\begingroup$ Not sure if I already posted this but I said "oops I was thinking of subgroups of order 3" (Android app is malfunctioning). $\endgroup$ – Matt Samuel Dec 11 '15 at 1:57
  • $\begingroup$ There is an alternative simple combinatorial argument that I've posted on this site before. If you're allowed to use Cauchy's theorem, you can deduce that all elements must be of the form $x^ay^b$, where $x$ and $y$ are elements of order 2 and 3. Then try to determine which element $yx$ it's. $\endgroup$ – Matt Samuel Dec 11 '15 at 2:07
  • $\begingroup$ @HenningMakholm I agree, this does not appear to be a proof. $\endgroup$ – Slade Dec 11 '15 at 2:35
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In the first part of the indented quote, you conclude that the inverse of an element of order 3 will itself has order 3, so order-3 elements come in pairs. So far so good. But the jump from there to "Then $a_1^2=b_1$, $a_2^2=b_2$, $a_3^2=b_1$, or $a_3^2=b_2$" is completely unjustified. In fact at this point you haven't really used the fact that $G$ has only 6 elements -- would however argument you have in mind there work as well to show that $S_4$ cannot have three different elements of order $3$ (which is manifestly untrue)? If it won't, then you need to explain how it works.

What I would say is:

We know that order-3 elements come in inverse pairs, and there is at least such pair, $a$ and $a^{-1}=a^2$. Suppose there is a different order-3 pair $b$ and $b^{-1}=b^2$; we then seek a contradiction.

What is the element $ab$? It cannot be $e$, because then $b$ would be $a^{-1}$, contradicting the assumption that $b$ is different from both $a$ and $a^{-1}$. It cannot equal $a$ or $b$ either because, say $a=ab$ implies $b=e$ by cancellation, but $b$ is assumed to have order $3$. And it cannot equal $a^2$ or $b^2$, because, say, $a^2=ab$ implies $a=b$ which was assumed not to be the case.

So $ab$ must be the sixth element of the group. And likewise $ab^{-1}$ must also be that sixth element of the group, and so we have $ab=ab^{-1}$. But then $b=b^{-1}$, so $b^2=e$, so $b$ didn't have order 3 after all, a contradiction!

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  • $\begingroup$ This is a very nice elementary approach to this problem. $\endgroup$ – David Hill Dec 11 '15 at 4:35
  • $\begingroup$ I assumed that $G$ has three elements of order 3, and hence my argument which you cited. $\endgroup$ – sequence Dec 11 '15 at 17:00
  • $\begingroup$ @sequence: How does that assumption lead to your conclusion? How does your supposed argument NOT work for $S_4$ which has plenty more than three elements of order 3? $\endgroup$ – Henning Makholm Dec 11 '15 at 17:01
  • $\begingroup$ @HenningMakholm: I used the fact that there are only 6 elements in $G$ (implicitly). But I didn't consider the possibility of $G$ having four elements of order 3. $\endgroup$ – sequence Dec 11 '15 at 17:05
  • $\begingroup$ @sequence: How do you use the fact that there are only 6 elements in $G$ to reach your conclusion? That's what you're supposed to explain when you promise a proof -- not just assert the conclusion while keeping to yourself what the argument you're envisaging is. Your alternative plan will work too, if you actually state that plan instead of leaving it to the reader to guess it. $\endgroup$ – Henning Makholm Dec 11 '15 at 17:09
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$G$ has an element $h$ of order $3$ by Cauchy's theorem. Then $H = \langle h \rangle$ is a subgroup of order $3$ containing two elements of order $3$, namely $h$ and $h^{-1}$.

If $G$ has some other element $k$ of order $3$, then $K = \langle k \rangle$ and $H = \langle h \rangle$ are distinct subgroups of order $3$. Now $H \cap K$ must be trivial, since its order must divide $|H| = |K| = 3$ and cannot be $3$ (otherwise $H$ and $K$ are equal). Therefore: $$|HK| = \frac{|H||K|}{|H \cap K|} = \frac{3\cdot 3}{1} = 9$$ But this is absurd because $G$ only has six elements.

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  • $\begingroup$ Cauchy's theorem is too advanced here. If there is no element of order 3, then every element is of order 2 (by Lagrange). It is now easy to conclude that $G$ is abelian. $\endgroup$ – David Hill Dec 11 '15 at 4:33
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    $\begingroup$ @DavidHill: The OP mentioned Cauchy's theorem in his question, so I assumed it was available. $\endgroup$ – Bungo Dec 11 '15 at 4:36
  • $\begingroup$ If you feel comfortable using a bazooka to kill a mosquito, then OK. $\endgroup$ – David Hill Dec 11 '15 at 4:39
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    $\begingroup$ I do feel comfortable with it, Cauchy's theorem is pretty basic. In any case, the OP already used it to establish that there are at least two elements of order $3$. I was just summarizing that before adressing the question in posed the title, which is why can't $G$ have more than two elements of order $3$. We don't need Cauchy's theorem for that part. $\endgroup$ – Bungo Dec 11 '15 at 4:47
  • $\begingroup$ The group has order 6. $\endgroup$ – David Hill Dec 11 '15 at 5:14
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Suppose $G$ has order $6$, with identity element $e$. By Cauchy's Theorem $G$ has at least one element of order $2$ and at least two elements of order $3$, because if $a \in G$ has order $3$, then so does $a^2$.

Let $b \in G$ be an element of order $2$. Then $G$ is generated by $a$ and $b$.

We have a subgroup $A=\{1,a,a^2\}$ in $G$ of index $2$, so $Ab=bA$ and $\{b,ab,a^2 b\}=\{b,ba,ba^2 \}$ and hence we have $ab=ba^2$ or $ab=ba$. In the first case, $(ab)^2=abab=ba^2 ab=b^2=1$ (order 2) and $a^2ba^2 b = a^2 abb=1$ (order 2), so done.

In the second case $G$ is abelian and $G \cong \mathbb{Z}_{6}$ ($abab=a^2$, so $ab$ has order $6$), so done. $\square$

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    $\begingroup$ This is essentially the proof I was talking about in my comment. You got my upvote for that, but I almost didn't bother reading it because it looks like an ugly dense block of text. Try using paragraphs. $\endgroup$ – Matt Samuel Dec 11 '15 at 4:36
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    $\begingroup$ Cauchy's theorem is too advanced. By Lagrange's theorem, every element has order either 2 or 3. If every element has order 2, then $G$ is abelian (easy to prove). Otherwise, the proof is good. @MattSamuel same comment for you. $\endgroup$ – David Hill Dec 11 '15 at 4:38
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    $\begingroup$ As far as I know, Largrange's theorem says that it is possible for a group of order 6 to have subgroups (or elements) of orders 2 and 3, but it's not necessarily the case that there are such elements (or subgroups). Thus I think Cauchy's theorem is a better argument in this case. $\endgroup$ – sequence Dec 11 '15 at 17:13

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