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Can one show that there must be only two groups of order 6 and that every abelian group of order 6 must be isomorphic to $C_6$ without doing an entire classification of groups (i.e. showing from the start that all groups of order 6 must be isomorphic to either $C_6$ or $S_3$)?

The context of this query is that I want to easily show that $\operatorname{GL}_2(\mathbf{F}_2)$ is isomorphic to $S_3$ without doing the whole work of classifying groups of order $6$ (and so the with the stategy above all that remains is to show that $\operatorname{GL}_2(\mathbf{F}_2)$ and $S_3$ are non-abelian and hence they must be isomorphic with the information above).

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  • $\begingroup$ I don't really understand your point. It seems like what you claim to want to do is to classify groups of order $6$ without... classifying groups of order $6$. It doesn't really make sense. But you can certainly show that the two groups are isomorphic without doing the classification. $\endgroup$ – tomasz Mar 25 at 21:52
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Suppose $G$ is any group with $|G| = 6$. There exist $a\in G$ of order $2$ and $b\in G$ of order $3$ by Cauchy's theorem. If $G$ is abelian, then $ab$ has order $6$, and so $G \cong C_6$. Otherwise, the subgroup $N = \langle b \rangle$ is normal because it has index $2$, and $N$ and $H = \langle a \rangle$ have trivial intersection and satisfy $NH = G$, so $G$ is a semidirect product of $N$ and $H$, which is defined by a nontrivial (since $G$ is not abelian) homomorphism $\phi: H \to \text{Aut}(N) \cong C_2$, of which there is only one. So there are only two groups of order $6$, and only one is abelian, namely $C_6$.


Here's an alternative, more elementary approach. I use Lagrange's theorem, but that's certainly something you'll learn very soon if you haven't already. I'll show there's only one nonabelian group of order $6$, since that seems to be what you really need. (This ends up basically being a proof of the classification, so maybe it's not so great.)

Suppose $G$ is a non-abelian group with $|G| = 6$. By Lagrange's theorem, the order of any element of $G$ is $1$,$2$,$3$, or $6$. If there is an element of order $6$, then $G$ is cyclic and thus abelian, so there is no element of $G$ with order $6$. If every nonidentity element of $G$ has order $2$, then if $a,b\in G$, we have $abab = (ab)^2 = 1$, and thus, multiplying both sides on the right by $ba$, we have $ab = ba$, so $G$ is abelian, and thus there must be some element of $G$ with order $3$. If every nonidentity element has order $3$, then letting $a \in G$ have order $3$, let $b\in G \setminus \{1,a,a^2\}$, and so $b^2 \notin \{1,a,a^2\}$, since otherwise $(b^2)^2 = b$ would be in this set, and it is not. Now let $c \in G \setminus \{1,a,a^2,b,b^2\}$. By similar reasoning, $c^2 \notin \{1,a,a^2,b,b^2\}$, but also $c^2 \ne c$ since $c$ has order $3$. Therefore $G$ contains at least $7$ elements, which is not the case. So there must be some element of $G$ with order $2$.

Let $a\in G$ have order $2$ and $b\in G$ have order $3$ (possible by the above paragraph). You can verify easily that $G = \{1, a, b, b^2, ab, ab^2\}$, as any equality between two of these elements leads quickly to a contradiction (of the facts that $a$ has order $2$ and $b,b^2$ have order $3$), and there are six of them. You can also check that $aba$ is equal to either $b$ or $b^2$ by eliminating the other possibilities in a similar way (equate $aba$ to the other elements in $\{1, a, b, b^2, ab, ab^2\}$ and reach contradictions easily). If $aba = b$, then by multiplying on the right by $a$, we get $ab = ba$, and so $G$ is abelian since $a$ and $b$ generate $G$ and commute. Therefore $aba = b^2$, which implies $ba = ab^2$. You can use this rule to reduce any string of $a$s and $b$s to the form $a^ib^j$ with $0\le i \le 1$ and $0 \le j \le 2$, which is the form in our list $G = \{1, a, b, b^2, ab, ab^2\}$, and so this rule determines the multiplication rule on $G$. Since $a = (12)$ and $b=(123)$ in $S_3$ satisfy these rules, $G$ is isomorphic to $S_3$.

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  • $\begingroup$ That seems like complex machinery that we didn't cover in our course yet, but anyway thank you very much for this answer (I upvoted you). I wonder if there's a simpler answer here otherwise I will only have to do the entire classification approach. $\endgroup$ – aaaaaaaah Mar 25 at 18:25
  • $\begingroup$ Is pretty much all of what I wrote using machinery you don't have? I'm happy to look for a different approach. $\endgroup$ – csprun Mar 25 at 18:29
  • $\begingroup$ I mean we didn't even cover Cauchy's theorem xD Plus stuff like "trivial intersection" and "semidirect product" is still in the unknown territory for me. $\endgroup$ – aaaaaaaah Mar 25 at 18:30
  • $\begingroup$ Okay, I'm adding an elementary approach to my answer. $\endgroup$ – csprun Mar 25 at 18:38
  • $\begingroup$ Let me know if my addition is readable with the machinery you currently have. $\endgroup$ – csprun Mar 25 at 19:02
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Rather than resorting to classification (which you do seem to try to do, despite claiming to want to do otherwise), you can actually explicitly define an isomorphism between $\operatorname{GL}_2(\mathbf F_2)$ and $S_3$.

Consider the set $A=\{(1,0),(0,1),(1,1)\}$ of vectors in plane over the two-element field. $S_3$ is clearly isomorphic to the group $\operatorname{Sym}(A)$ of all permutations of $A$. You can show that the natural action of $\operatorname{GL}_2(\mathbf F_2)$ on the plane preserves $A$. This shows that the restriction of $\operatorname{GL}_2(\mathbf F_2)$ to $A$ is a group action, which yields a homomorphism $\phi\colon \operatorname{GL}_2(\mathbf F_2)\to \operatorname{Sym}(A)\cong S_3$. Then you can use any two of the three following observations (all fairly easy) to show that it is an isomorphism:

  1. $\phi$ is injective,
  2. $\phi$ is surjective,
  3. the two groups have the same number of elements (and are finite), so $\phi$ is injective if and only if it is surjective.

Alternatively, you can define the inverse of $\phi$. This is fairly easy, but proving that it is a well-defined isomorphism won't be quite so elegant, I think.

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