An inequality for the minimal number of generators of a finite group II Let $G$ be a finite group, $n(G)$ the minimal number of generators and $m(G)$ the minimal number of irreducible complex representations generating exactly the left regular representation (with $\otimes$, $\oplus$).    
Question: Is it true that $n(G) \ge m(G)$ ?
 A: The inequality seems to be false for suitably large dihedral groups.  Indeed, if $G$ is any dihedral group, then $n(G)=2$.  It suffices to check that $m(G)>2$ for $G$ big enough.
Take $G=D_{29}$, the dihedral group of order $58$.  Its character table can be found here; it admits $2$ one-dimensional and $14$ two-dimensional irreducible representations.  Denote them by $A_1, A_2$ and $E_1, \ldots, E_{14}$, respectively.  We will show that any two of these representations are not sufficient to generate the regular one by using $\oplus$ and $\otimes$. In this situation, the regular representation is isomorphic to 
$$ A_1\oplus A_2 \oplus \bigoplus_{i=1}^{14}E_i^{\oplus 2}.$$ 
The above link also contains the tensor product table for $D_{29}$.  It can be completely described by the following rules:


*

*$A_1\otimes V = V$ for any $V$;

*$A_2\otimes A_2 = A_1$;

*$A_i\otimes E_j = E_j$ for any $i$ and $j$;

*$E_i\otimes E_i = A_1\oplus A_2\oplus E_{2i}$;

*$E_i\otimes E_j = E_{|i-j|}\oplus E_{i+j}$,


where we put $E_{15}=E_{14}, E_{16}= E_{13}, \ldots, E_{28}= E_{1}$.
Obviously $A_1$ and $A_2$ alone cannot generate the regular representation, so suppose first that the regular representation can be obtained from $A_i$ and $E_j$ for some $i$ and $j$.  Since $A_i\otimes E_j = E_j$,  $E_j\otimes E_j= A_1\oplus A_2\oplus E_{2i}$ and $E_j\otimes E_j \otimes E_j = E_j\oplus E_j\oplus E_j\oplus E_k$ for some $k$, we have that $E_j^{\otimes 3}$ cannot be part of the regular representation (since it contains three copies of $E_j$), so we can only use $A_i, E_j$ and $E_j^{\otimes 2}$, which is clearly not enough to generate the regular representation.  Thus $A_i$ and $E_j$ cannot generate it.
Now, suppose that $E_i$ and $E_j$ generate the regular representation for some $i$ and $j$.  Again, $E_i^{\otimes 3}$ and $E_j^{\otimes 3}$ cannot be part of it, for the same reason as above.  Thus we can only use $E_i, E_j, E_i^{\otimes 2}, E_j^{\otimes 2}, E_i\otimes E_j, E_i^{\otimes 2}\otimes E_j$ and $E_i\otimes E_j^{\otimes 2}$.  Also, they can appear at most $2,2,1,1,2,1$ and $1$ times, respectively, since the $A_i$ appear only once and the $E_k$ twice each in the regular representation.  Thus the largest representation we can obtain by using these has dimension $4+4+4+4+8+8+8 = 40$.  Since the regular representation has dimension $58$, it cannot be generated by $E_i$ and $E_j$.
The last two paragraphs show that we need more than two irreducible representations to obtain the regular one. Hence $m(D_{29})>2$, even though $n(D_{29})=2$.
