I have this approach to classify groups of order $6$ and $10$. Let $G$ be a group with $|G|=n$
If $n=6$, then by Sylow there is an element $g$ of order $3$. Hence $N:=\langle g\rangle$ has index $2$ and is normal. Let $x \in G-N$ such that $G=N \cup xN$. Then $x^2\in N$ (otherwise $x^2 \in xN \Rightarrow x \in N $). If $x^2 \neq e$, then $\operatorname{ord}(x)=6$ since $x^3 \in xN$ and we find the cyclic group. If $x^2=e$, then the consider conjugation map $\gamma_x :N\longrightarrow N$ with $\gamma_x(g)=xgx^{-1}$. Now the step I'm going to ask about. Since $N=\{e,g,g^2\}$ it's either $g=xgx^{-1}$ or $g^2=xgx^{-1}$. In the first case $G$ is abelian (and cyclic by the Chinese remainder theorem) and in the second case isomorphic to the dihedral group.
What if $n=10$? I can proceed exactly as in the case $n=6$, but the conjugation map could now have many different images: i) $\gamma_x(g)=xgx^{-1}=g$ (in this case $G$ is abelian) ii) $\gamma_x(g)=xgx^{-1}=g^2$ iii) $\gamma_x(g)=xgx^{-1}=g^3$ iv) $\gamma_x(g)=xgx^{-1}=g^4$ (in this case $G\cong D_5$)
I don't see how one can lead ii) and iii) to any contradiction.
The contradiction is as follows: as $x^2=e$, then $\gamma_{x^2}=\gamma_e$ should be identity. However, it's not:
$$g=ege=x^2gx^{-2}=x(xgx^{-1})x^{-1}=xg^2x^{-1}=xg(x^{-1}x)gx^{-1}=(xgx^{-1})(xgx^{-1})=g^2g^2=g^4\ne g$$
(in case (ii), and similarly in case (iii)).