Determine all generators of a group Exersice:

Determine all generators of the group $\langle \mathbb
 Z_{11}^{*};\odot\rangle$

Their solution:

By the Lagrange's Theorem, it follows that for any $a\in \mathbb
 Z_{11}^{*}$ we have $ord(a)\in\{1,2,5,10\}$. If $a$ is a generator of
  $\mathbb Z_{11}^{*}$, then $a$ has the order $10=\varphi(11)$. This happens
  iff $a^2\neq 1$ and $a^5 \neq 1$. By trying all the possibilities, we
  get that 2,6,7,8 are the generators of $\mathbb Z_{11}^{*}$

If see, that we get $ord(a)\in\{1,2,5,10\}$. I don't know why, it's basically just a theorem I apply, can maybe someone give me a little explanation to Lagrange's Theorem?
Anyway, now we know the possible orders of a. From Fermat/Euler we know that 

$\forall m\geq 2$ and for all $a$ with $gcd(a,m)=1: a^{\varphi(m)}=_m 1$

Which someone seems to be used here. What they basically do is, check $a^2\neq 1$ and $a^5\neq 1$ for all $a\in [1,10]$. And the ones that produce a 1, aren't generators - the others are.
But I can't really follow why. I mean, I see, why we don't check 1 and 10 since $a^10=e=1$ [I knwo that from some colloray, but no idea why exactly]. And $a^1 = a$, so we don't have to check that either.
So, can someone give me the basic idea behind this? I did quit a bit of research but there's just some "problem" in my head I can't get fixed.
Basically, I don't get Euler/Fermat and Lagrange. 
 A: Lagrange's Theorem states that if $G$ is a finite group and $H$ is a subgroup of $G$, then $|H|$ divides $|G|$. In particular, for $a \in G$, $\langle a \rangle$ is a subgroup of $G$, so $|a| = |\langle a \rangle|$ divides $G$. 
Another consequence of Lagrange's Theorem is that for $a \in G$, $a^{|G|} = 1$. Take $q$ to be the order of $a$ in $G$. So $|G| = qk$, for some integer $k$. Thus, $a^{|G|} = (a^{q})^{k} = 1^{k} = 1$. 
Now $11$ is prime, so $\mathbb{Z}_{11}^{\times} \cong \mathbb{Z}_{10}$ (we note that $\phi(10) = 11$). So a generator of $\mathbb{Z}_{11}^{\times}$ must have order $10$ in the multiplicative group. What Lagrange's Theorem tells us is that every element in $\mathbb{Z}_{11}^{\times}$ has order dividing $10$ (since $|\mathbb{Z}_{11}^{\times}| = 10$). So every element has order $1, 2, 5$, or $10$. 
Now the Euler-Fermat Theorem is a corollary of Lagrange's Theorem. The multiplicative group $\mathbb{Z}_{n}^{\times}$ has order $\phi(n)$. So for any element $a \in \mathbb{Z}_{n}^{\times}$, $a^{\phi(n)} \equiv 1 \pmod{n}$. 
