Partition groups into subcollections of isomorphic groups - Fraleigh p. 84 8.10 (please revamp title if necessary) Here a * superscript means
all nonzero elements of the set. The orange is the answer. 

Then S = $\{C_1, ..., C_9\}$ is a partition of the given
collection into subcollections of isomorphic groups.
Notice that C1 consists of all the infinite cyclic groups;
C2 consists of the cyclic groups of order 2;
C3 consists of the cyclic groups of order 6;
C4's groups are isomorphic due to $ln : \langle \mathbb{R}^+, \cdot \rangle \rightarrow  \langle \mathbb{R}, \cdot \rangle$
 C5 contains the only non-abelian group in the collection;  
C6 isn't cyclic, and is countably infinite. but notice $\langle \mathbb{Q}*, \cdot \rangle$ is not isomorphic to
$\langle \mathbb{Q}, + \rangle$ since 
$x*x = id \implies \begin{cases} x \cdot x = 1 \text{ has two solutions in } \langle \mathbb{Q}*, \cdot \rangle \\ x + x = 0, \text{has one solution in } \langle \mathbb{Q}, + \rangle \end{cases}$; 
(C7.) the solutions of the equation $x^2 = 1$ in $\langle \mathbb{R}^*, \cdot \rangle$ and
$\langle \mathbb{R}^+, \cdot \rangle$ show these groups are non-isomorphic.
$x*x = 1 \implies \begin{cases} x \cdot x = 1 \text{ has two solutions in } \langle \mathbb{R}^*, \cdot \rangle \\ x + x = 1, \text{has one solution in } \langle \mathbb{R}^+, \cdot \rangle \end{cases}$
(C8.) Finally the equation $x^4 = 1$ has four solutions in $\mathbb{C}$. This distinguishes $\langle \mathbb{C}*, \cdot \rangle$ from all the other
infinite groups in the collection. I know $x^4 = 1 \implies x = \pm 1, \pm i$. 

(1.) What's a smart efficient way to do this? I know to determine if two groups are isomorphic, then a first bid is to use cardinality. If they don't have the same cardinality, they can't be isomorphic. Then what?
(2.) (By dint of fkraiem's answer) How and why is it true for any two groups:
  "The different number of solutions to any equation involving the identity (like $x*x = \mathrm{id}$) means that if you try to construct an isomorphism, you will run into a contradiction."
I want to use this in general, hence do you need to prove this? Or is there intuition?
  I see groups C6, C7 differ in their number of solutions to $x * x = id$. 
(3.) The trick to fkraiem's answer is to write $\phi(1) = 0$ as $\phi((-1)\times(-1))$. Does this work on the whole? How do you predestine this trick? 

 A: (2.) The different number of solutions to $x*x = \mathrm{id}$ (or some other equation involving the identity) means that if you try to construct an isomorphism, you will run into a contradiction. Say I have an isomorphism $\phi: (\mathbf{Q}^*,\times) \to (\mathbf{Q}, +)$. Since it is an isomorphism, I have $\color{brown}{\phi(1) = 0}$.
I write $\phi((-1)\times(-1))$  in two ways.
First, $\phi((-1)\times(-1)) = \color{brown}{\phi(1) = 0}. \quad (♥)$
Secoond, $\phi((-1)\times(-1)) = \phi(-1)+\phi(-1) = 2\phi(-1)$.
By dint of $(♥)$, this $= 0$. Hence $2\phi(-1) = 0$,
contradicting the fact that $\phi$ is an isomorphism and is thus injective.
EDIT: In more group-theoretic terms, $(\mathbf{Q},+)$ has no element of order two, while $(\mathbf{Q}^*,\times)$ has one, so they can't be isomorphic, since two isomorphic groups have the same group structure.
EDIT after OP's edit to his question:
The fact of the matter is simply what I said in my first edit: two groups are not isomorphic if there is a structural difference between them.
In this case, $\mathbf{Q}$ has no element of order two (there is no non-zero rational $x$ such that $2x = 0$), while $\mathbf{Q}^*$ has one (there is one non-one rational $x$ such that $x^2 = 1$, namely $-1$). If $\phi: \mathbf{Q}^* \to \mathbf{Q}$ is an isomorphism, then for any $x \in \mathbf{Q}^*$ of order two, $\phi(x) \in \mathbf{Q}$ has order two. So $\mathbf{Q}$ has at least one elements of order two, which is absurd since it has none.
And, again, this is only one kind of "structural difference", there are others. For example, you can show that $\mathbf{R}^*$ and $\mathbf{C}^*$ are not isomorphic by using the fact that every element of $\mathbf{C}^*$ has a square root, which is not true in $\mathbf{R}^*$. (If $z \in \mathbf{C}^*$ has a square root $a$, then $\phi(z) \in \mathbf{R}^*$ has square root $\phi(a)$...)
