Show $Dic_{3} \not \cong D_{6}$
def $$ Dic_{3} = \langle a_1,d_1 : a_1^6= e , d_1^2 = a_1^3,d_1a_1 =a_1^{-1} d\rangle $$
and $$ D_6 = \langle a_2,d_2: a_2^6=e,a_2^2 =(a_2d_2)^2=e \rangle $$
what is the easiest way to prove they are not isomorphic. I am tempted to make tables for both and have that they are not the same structure from there. Another way is to draw the sub lattice diagrams of each. But seems like work for a computer. Might ask questions for tables separately.
Guessing the elegant way would be to reach contradiction from the rules where there are two possibility either $a_1=a_2$ and $d_1=d_2$ or $a_1=d_2$ and $d_1=a_2$
case i $a_1=a_2$ and $d_1=d_2$
the order of a is 6 in dic_3 but then it is 2 in $d_6$
[contradiction]
case ii
It is obvious there is contradiction
( actually I do not know what is obvious. I am sure it might take a paragraph to justify)
[contradiction]
Easiest way to distinguish them is (in my opinion) counting homomorphisms to $\Bbb Z/2$. They are obviously determined by image of generators. Let's see, if $t$ is any homomorphism from $Dic_3$, then $t(d^2) = t(d)^2 = 0 = t(a^3)$; 2 and 3 are coprime, so $t(a) = 0$ and we have two homomorphisms — trivial and $(d \mapsto 1, a \mapsto 0)$.
For $D_6$ you can see that any map of generators to $\Bbb Z/2$ gives homomorphism (because all relations of dihedral group are laws in $\Bbb Z/2$) and there are 4 of them.