Difference between $\forall a. \exists b. c$ and $\exists b. \forall a. c$ Let's say we talk about groups ($A$ is the group) and inverse elements (called $i$):
$\forall a \in A. \exists i \in A. a \ast i = e$, and $\exists i \in A.\forall a \in A. a \ast i = e$ are obviously different. 
The first one says, the inverse element is dependent on the element taken from the group, which is correct. Whereas the second says there is the same inverse element for each element taken from the group, which is not (necessarily) correct.
But when we talk about idenitity elements:
$\forall a \in A. \exists e \in A. a \ast e = a$, and $\exists e \in A.\forall a \in A. a \ast e = a$ are of course still different, but it seems to me that both are correct.
This leads me to believe that $\exists b. \forall a. c$ is a stronger claim than $\forall a. \exists b. c$, but the first one induces the second one.
Is this assessment correct?
 A: Your assumption is correct. $\exists a \forall b \phi(x,y)$ (where $\phi(x,y)$ is some formula with occurences of $a$ and $b$) is a stronger claim than $\forall b \exists a \phi(a,b)$ in the sense that (1) logically implies (2), but not vice versa.  
You can verify this by considering the truth conditions of the two forumlas:  


*

*(1) $\Longrightarrow$ (2):
$\exists a \forall b \phi(a,b)$ is true under an assignment iff there exists an element $m$ from the domain for which $\forall b \phi(\widetilde{m},b)$ is true.*  $\forall b \phi(\widetilde{m},b)$ is true iff for all elements $m'$ from the domain, $\phi(\widetilde{m},\widetilde{m'})$ holds. So (1) is true iff there is an $m$ such that for all $m'$, $\phi(\widetilde{m},\widetilde{m'})$ is true. But then, if there exists a single $m$ which works for all $m'$, obviously we will also find such an $m$ for any $m'$. I.e., for all $m'$ it holds that $\exists a \phi(a,\widetilde{m'})$, and hence, $\forall b \exists a \phi(a,b)$ is true.  

*(2) $\not \Longrightarrow$ (1):
Proving $\forall b \exists a \phi(a,b)$ amounts to showing that for any element $m'$, we will find an $m$ such that $\phi(\widetilde{m},\widetilde{m'})$ is true. But this does not imply that there is a unique $m$ that works for all $m'$ (as would be the case in formula (1)): We might need to choose different $m$'s for any of the $m'$  we consider; then statement (2) is true, but (1) would be false, since there is no $m$ which makes the statement true for any $m'$. Thus (2) does not logically imply (1), but doesn't contradict it either, and is hence a weaker statement. 


Applied to your example, as you correctly figured out, while it is true that for any $a$ we will find some $i$ such that $a*i=e$, this does not guarantee that we will find a unique $i$ such that $a*i=e$ becomes true for any $a$. Hence, $\forall a \in A. \exists i \in A. a \ast i = e$ does not logically imply  $\exists i \in A.\forall a \in A. a \ast i = e$.
Conversely, if we did find such an $i$ that satisfies $a*i=e$ for any $a$, it would immediately follow that there is such an $i$ for any $a$. Hence, if $\exists i \in A. \forall a \in A. a \ast i = e$, then $\forall a \in A. \exists i \in A. a \ast i = e$ logically follows.

* I'm using the lazy notation $P(\widetilde{m})$ here to denote that the denotation of the predicate $P$ holds for the individual $m$ - be aware that mixing elements of the domain (i.e. real world objects) into formulas (i.e. a string of symbols) like $P(m)$ is not correct syntactically, but anything else makes the argument rather clumsy.  
