I think I just didn't get the core of group theory. Although it makes sense to me to follow the regular steps to solve problems of group theory.
For example, a group of order $10$ is isomorphic to $\mathbb{Z}_{10}$. To prove this, the standard solution suggests that we have to suppose that there are $2$ elements $x,y$. $x$ has order of 5 another has order of $2$ to begin with. And finally applied that it's isomorphic to $\mathbb{Z}_{2}\times\mathbb{Z}_{5}$ and then isomorphic to $\mathbb{Z}_{10}$
I am confused that, isn't that obvious that a group of order 10 is isomorphic to Z10??
Both of them have $10$ elements. We can simply project them one-by-one..
like $1$ to $x$ ; $2$ to $e$ ; $3$ to $y$….
…….
…….
Best Answer
It is false that a group of order $10$ must be cyclic.
There are two isomorphism types of groups of order $10: an abelian group, which is indeed cyclic, and a nonabelian group.
The nonabelian group of order $10$ is the dihedral group of degree $5$ (you may see it denoted as either $D_5$ or $D_{10}$). It can be realized as the group of rigid motions of a regular pentagon. It has presentation: $$\Bigl\langle r,s\;\Bigm|\; r^5 = s^2 = 1, sr=r^{4}s\Bigr\rangle.$$
But with your final paragraph: it is not enough for them to have the same number of elements. For example, $\mathbf{Z}_2\times\mathbf{Z}_2$ and $\mathbf{Z}_4$ both have 4 elements, are both abelian, but they are not isomorphic, because the latter group has only two solutions to $x+x=0$ (namely, $x=0$ and $x=2$), but the former group has four solutions to that equation (every element is a solution). So you don't know ahead of time that every group of order $10$ (or even that every abelian group of order $10$) must be cyclic.