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?

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Note: In Doerk's it says it's a simplified version of Schreier Subgroup Theorem.

  • $\begingroup$ did you really copy it verbatim? $\endgroup$ – Jorge Jan 2 '17 at 18:12
  • $\begingroup$ I'm confused: what "original proposition" is that? In Doerk's book or in Hubbert's? $\endgroup$ – DonAntonio Jan 2 '17 at 18:12
  • $\begingroup$ Original proposition is Hubbert's. And yes, I copied it verbatim. $\endgroup$ – LeviathanTheEsper Jan 2 '17 at 18:13
  • $\begingroup$ Is there a reason to assume the elements of the generating set must be distinct? Since there are no assumptions on the generating set $\{g_1,\ldots,g_n\}$ of $G$, it seems to me like the goal of this theorem may just be to give an upper bound on the size of a generating set of $U$. $\endgroup$ – sTertooy Jan 2 '17 at 18:15
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    $\begingroup$ The second sentence translates as "Then $U$ can be generated with $2n|G:U|$ elements." That does not imply that these elements are distinct. The word "set" is not used. $\endgroup$ – Derek Holt Jan 2 '17 at 18:49

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.

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    $\begingroup$ PS: I really don't like the fact tat he uses big cap letters to represent elements of the group. $\endgroup$ – Jorge Jan 2 '17 at 18:39

The proposition does not say that the elements need to be distinct.

  • $\begingroup$ The proposition says that the generating set has $2n|G:U|$ elements. In the way it's stated, they should be distinct. Perhaps I misinterpreted it that way. $\endgroup$ – LeviathanTheEsper Jan 2 '17 at 18:27
  • $\begingroup$ @LeviathanTheEsper Look at the proof. Where does he show that the elements $G_{i_1},$ etc are distinct? $\endgroup$ – Igor Rivin Jan 2 '17 at 18:32
  • $\begingroup$ And as far as I can see, Huppert does not use the word set (which would be Menge in German). $\endgroup$ – Derek Holt Jan 2 '17 at 18:40

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