Why group of order 6 has to have just two elements of order 3

Question

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 $G$.

Answer 1

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!

Answer 2

$G$ 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.

Answer 3

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. $\square$