# 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…

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).

• 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. – tomasz Mar 25 at 21:52

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 $$S_3$$.

• 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. – aaaaaaaah Mar 25 at 18:25
• Is pretty much all of what I wrote using machinery you don't have? I'm happy to look for a different approach. – csprun Mar 25 at 18:29
• 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. – aaaaaaaah Mar 25 at 18:30
• Okay, I'm adding an elementary approach to my answer. – csprun Mar 25 at 18:38
• Let me know if my addition is readable with the machinery you currently have. – csprun Mar 25 at 19:02

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.