# Find all groups of order $6$.

Find all groups of order $$6$$.

If there is element of order $$6$$ then group is cyclic. If not, I came to conclusion that there should be an element of group $$G$$ of order $$3$$ (otherwise the order of group is $$2^{n},$$ for some $$n \in \mathbb{N}$$). Let that element be $$a$$. Now, let $$b \in G \setminus\langle a\rangle$$. It is easy to show that all elements $$e, a, a^{2}, b, ab, a^{2}b$$ are different. Now I am stuck. I am sure that this isn't enough information about group $$G$$ but since I am a beginner in this scope I don't know what else should I write. Any hint helps!

• Having the multiplication table is quite enough. Aug 3 '19 at 9:27
• There is only 1 option now for G: the non abeilian dihedral group $D_6$. Aug 3 '19 at 9:29
• @James True, but the OP is not supposed to rely on this knowledge at this point. Aug 3 '19 at 9:34
• @IvanNeretin good point, but I added it so that by looking at the multiplication table of $D_6$ they can see what they're heading for. Also this should help: math.stackexchange.com/questions/742975/… Aug 3 '19 at 9:42

Let us consider the case that there is no element of order $$6$$. We can find an element $$a \in G$$ of order $$2$$ and an element $$b \in G$$ of order $$3$$ (for example by Cauchy's theorem or by your argument). Thus we have the subgroups $$K = \langle a \rangle = \lbrace 1,a \rbrace$$ and $$H = \langle b \rangle = \lbrace 1,b,b^2 \rbrace$$. By Lagrange's theorem we get $$H \cap K = \lbrace 1 \rbrace$$. Suppose that $$ab = ba$$. Then one can show that the order of $$ab \in G$$ is $$6$$, a contradiction to our assumption. We get $$ab \neq ba$$, which yields $$G = HK$$ as you were basically stating as well. Thus $$G$$ is a semidirect product, namely $$G = H \rtimes K \cong C_3 \rtimes C_2 = D_3 \cong S_3$$.
I should have taken $$a$$ to be the element of order $$3$$ as you did. Do not get confused by that. My $$a$$ is your $$b$$ and vice versa.