If $G=\langle g_1,\dots, g_n\rangle$ and $U\leq G$ of finite index, then $U$ has a generating set with $2n|G:U|$ elements. What's wrong? I found this proposition in the Doerk's book of Finite Soluble Groups, which is referenced from the book Endliche Gruppen I from B. Huppert.

Let $G=\langle g_1,\dots, g_n\rangle$, and let $U$ be a subgrouup of $G$ of finite index. Then $U$ has a generating set with $2n|G:U|$ elements.

But if $G=D_4=\langle \alpha,\beta\mid \alpha^4=\beta^2=e,\beta\alpha=\alpha^3\beta\rangle$, and $U=\langle\alpha\rangle$, then $|G:U|=2$. But the proposition says that $U$ should have a generating set of $2\cdot 2\cdot 2=8>4=|U|$ elements which is impossible.
What's the problem?
By the way, he original proposition says

Sei $\mathfrak{G}=\langle G_1,\dots,G_n \rangle$ eine endlich erzeugbare Gruppe und $\mathfrak{U}$ eine Untergruppe von $\mathfrak{G}$ von endlichem Index. Dann ist $\mathfrak{U}$ mit $2n |\mathfrak{G}:\mathfrak{U}|$ Elementen erzeugbar.

which more or less means

Let $\mathfrak{G}=\langle G_1,\dots,G_n \rangle$ a finitely generated group and $\mathfrak{U}$ a subgroup of $\mathfrak{G}$ of finite index. Then $\mathfrak{U}$ has a generating set with $2n |\mathfrak{G}:\mathfrak{U}|$ elements.

It seems to me that the proposition is the same, am I missing something?
The proof can be found in the Hubbert's book, but I'm a novice in german and I don't understand, starting from Nebenklassenzerlegung. Is there any extra hypothesis used in the proof?

Note: In Doerk's it says it's a simplified version of Schreier Subgroup Theorem.
 A: The proposition does not say that the elements need to be distinct.
A: the way that the proof works is that it lets $G_1,G_2,\dots G_n$ be the generators of $\mathcal G$ and it makes $G_{n+j}=G_j^{-1}$, then it takes representatives $V_1,V_2,\dots V_m$ for the cosets of $\mathcal U$ in $\mathcal G$, such that $V_1$ is thentity of the group.
Now he defines $U_{i,j}$ as the solution to the equation $V_iG_j=U_{ij}V_k$, where $k$ is the representative for the appropriate coset (so that this $U_{i,j}$ is in $\mathcal U$).
We now prove that the $U_{i,j}$ generate $\mathcal U$.
Take an element of $\mathcal U$, then it is of the form $G_{a_1}G_{a_r}\dots G_{a_m}$ such that every $a_i\in \{1,2,3\dots,2n\}$.
We then have:
$G_{a_1}G_{a_2}\dots G_{a_r}=V_1G_{a_1}\dots G_{a_r}=U_{1,a_1}V_{a_1'}G_{a_2}\dots G_n=\dots U_{1,a_1}U_{a_1',a_2}\dots U_{a_{r-1}',a_r}V_{a_r'}$
and since all of the first factors are in $U$ we must have $a_r'=1$ and  so $U=U_{1,a_1}U_{a_1',a_2}\dots U_{a_{r-1}',a_r}$.
So the proof shows that the elements $U_{i,j}$ generate $\mathcal U$, but of course it doesn't show they are different, so we can build a genering set with at most $2mn$ elements.
