Possible group equation of a group of order $10$:
Consider $|G|=|Z(G)|+\sum _{i=1}^n |cl(a_i)|$ where $a_i's$ are class representatives.
If the group is abelian then $10=10$ is the only one.
If the group is non-abelian then the group is isomorphic to $D_5$.Hence class-equation will be done as follows:
$|Z(D_5)|=1$.
Also $|cl(a_i)|=\dfrac{|G|}{|C(a_i)|}$.For any element $x$; $C(x)=\{e,x,..\};|C(x)|\geq 2$.Also possible orders of $|C(x)|$ is $2,5,10$.If $|C(x)|=10$ then $x\in Z(D_5)$ contradiction. Thus $|C(x)|=2,5$
Now if $|C(x)|=2\forall x$ then $10=1+2+2+2+2+2$ which cant be hence $|C(x)|=5 $ for one $x$.
Hence the class equation becomes $10=1+5+2+2$ and it is the only one.
Am I correct ?Please help.
Best Answer
The solution is correct.
Other way: $G=$ non-abelian group of order $10$. Then $|Z(G)|<10$, and it can't be $2$ or $5$ (why?)
Then repeat same steps of your argument after you shown that center is trivial. They work, just because of $G$ is non-abelian (even tough we do not consider that it is actually $D_{10}$).