# If $G$ is solvable and $G/[G,G]$ is cyclic, can $G\times G$ be generated by 2 elements?

I doubt this is true, but I haven't found any small counterexamples (there are no counterexamples with $|G| < 1536, |G| \neq 768$):

Suppose $G$ is finite, solvable, 2-generated, and $G/[G,G]$ is cyclic. Must $G\times G$ be 2-generated?

Hall and Gaschütz have relevant papers ("eulerian functions" is probably in the title) that I have not had time to check.

This is motivated by A group generated by two elements such that its product with itself is not generated by two elements. If $G/[G,G]$ requires $k$ generators, then $G$ requires at least $k$ generators and $G \times G$ at least $2k$. In particular, if $G/[G,G]$ is not cyclic, then $G \times G$ is obviously not 2-generated, so we are mostly interested in $G$ with $G/[G,G]$ cyclic. Since groups with $G=[G,G]$ 2-generated, but $G\times G$ not 2-generated are plentiful but difficult to understand, we restrict to solvable groups.

• I guess it is worth mentioning that this fails for nonsolvable groups. By the classification, if $G$ is a finite simple group, then $G$ is generated by two elements. The minimal number of generators for $G^n$ (direct product of $n$ copies of $G$) goes to infinity as $n \rightarrow \infty$ (this is true for any nontrivial finite group). So for some $n$ we have $G^n$ generated by two elements, and $G^{n} \times G^{n}$ not generated by two elements (and the abelianization of $G^n$ is trivial cyclic). Commented Apr 15, 2014 at 19:04
• But I think that this might be true when $G$ is a finite simple group! At least most finite simple groups are generated by two elements $x$, $y$ such that $x$ has order $2$ and $y$ has order $3$ or $5$. Then $G \times G$ is generated by $(x,y)$ and $(y,x)$. Commented Apr 15, 2014 at 19:12
• Is it true that $G$ is generated by two elements if $G$ is solvable and $G/[G,G]$ is cyclic? Commented Apr 15, 2014 at 20:48
• No, $G$ need not be 2-generated (the generalized dihedral group on $C_3 \times C_3$ is a counterexample of minimal order, 18). I had forgotten to include that as a hypothesis. Commented Apr 16, 2014 at 3:00

This is not a complete answer, but I am starting to think that the answer might be yes, $G \times G$ is $2$-generated. (Added later: I believe that it is now a complete answer, and that the answer is indeed yes.)

I can prove it under the following additional assumption:

There exist $a \in G$, $b \in G'$ with $\langle a,b \rangle = G$.

Assuming that is true, let $G_1$ and $G_2$ be copies of $G$ with corresponding generators $a_1,b_1,a_2,b_2$, and let $H = \langle a_1b_2,a_2b_1 \rangle \le G_1 \times G_2$. Then I claim that $H=G_1 \times G_2$.

Clearly $H$ projects onto both direct factors, so $H$ is a subdirect product of $G_1 \times G_2$. So, if $N_i = H \cap G_i$ for $i=1,2$, then $N_i \unlhd G_i$ and $H/(N_1 \times N_2)$ is a diagonal subgroup of $G/(N_1 \times N_2)$.

Also, since $b_i \in G_i'$, we must have $G_i =\langle a_i, G_i' \rangle$, and so $H$ projects onto the factor group $G_1/G_1' \times G_2/G_2'$, and hence also onto $G_1/N_1G_1' \times G_2/N_2G_2'$. But if $N_i \ne G_i$ then, since $G/N_i$ is solvable, $G_i \ne N_iG_i'$, and then it is not possible for a diagonal subgroup of $G/(N_1 \times N_2)$ to project onto the whole of the direct product $G_1/N_1G_1' \times G_2/N_2G_2'$, so we must have $H=G_1 \times G_2$.

So is the extra assumption always satisfied? The answer is certainly yes if $G/G'$ has prime power order or is small. Since we can always replace the geenrator $a$ by $ab^i$ for any $i \in {\mathbb Z}$, it would be satisfied if the answer to the following question is yes:

If $a$ and $b$ generate the finite cyclic group $G$, then does $G$ necessarily have a single generator of the form $ab^i$ or $a^ib$ for some $i \in {\mathbb Z}$?

Added later: I claim now that if the cyclic group $C_n$ of order $n$ is generated by $a$ and $b$, then it is generated by the single element $ab^i$ for some $i \in {\mathbb Z}$, which completes the proof that the answer to the question is yes.

Let $n = q_1q_2 \cdots q_r$, where the $q_i$ are powers of distinct primes. Then there is an isomorphism $C_n \to C_{q_1} \times \cdots \times C_{q_r}$. Denote the images of $a$ and $b$ under this isomorphism be $(a_1,\ldots,a_r)$ and $(b_1,\ldots,b_r)$. The condition that $C_n = \langle a,b \rangle$ is equivalent to the condition that, for each $i$, at least one of $a_i$ and $b_i$ generates $C_{q_i}$.

Choose the numbering such that $a_i$ and $b_i$ both generate $C_{q_i}$ if and only if $1 \le i \le k$, and let $m = q_1q_2 \cdots q_k$ (where $m=1$ if $k=0$). Then, for $i > k$, $b_i$ generates $C_{q_i}$ if and only if $b_i^m$ does and so, for each $i$ with $1 \le i \le r$, exactly one of $a_i$ and $b_i^m$ generates $C_{q_i}$. So $a_ib_i^m$ generates $C_{q_i}$ for each $i$, and hence $ab^m$ generates $C_n$.

• Yes, corrected! Commented Apr 16, 2014 at 14:08