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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!

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    $\begingroup$ Having the multiplication table is quite enough. $\endgroup$ Aug 3 '19 at 9:27
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    $\begingroup$ There is only 1 option now for G: the non abeilian dihedral group $D_6$. $\endgroup$
    – James
    Aug 3 '19 at 9:29
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    $\begingroup$ @James True, but the OP is not supposed to rely on this knowledge at this point. $\endgroup$ Aug 3 '19 at 9:34
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    $\begingroup$ @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/… $\endgroup$
    – James
    Aug 3 '19 at 9:42
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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.

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