# Is $G=\left<b_1, b_2, b_3 | [b_i^p, b_j^p]=1, \forall i,j=1,2,3\right>$ large?

Dear all,

I came across the above group (for any fixed odd prime $p$) and would need to know if it's large or not. It seems like it shouldn't be...

Best wishes, Elisabeth

• What do you mean by "small"? I assume that $p$ is a fixed natural number. The group is certainly infinite because $b_1,b_2,b_3$ generate a copy of $\mathcal{Z}^3$. Whenever $p>1$, $G$ is not amenable because $b_1,b_2b_1b_2^{-1}$ generates a free group – Steven Deprez Apr 26 '12 at 15:20
• Maybe "large" in the sense of subjecting onto a free group? – Igor Rivin Apr 26 '12 at 15:25
• subjecting -> suRjecting. Thanks, Apple, for the spelling correction... – Igor Rivin Apr 26 '12 at 15:45
• @Igor: large means there's a finite-index subgroup surjecting a rank 2 free group. – Ian Agol Apr 26 '12 at 15:54
• Your group is obtained by adjoining roots to each generator of $\mathbb{Z}^3$. In general, if you add an $n$th root to an element $\gamma$ of a group $\Gamma$, then there is an index-$n$ subgroup isomorphic to an iterated $n$-fold amalgamated product of $\Gamma$ along $\gamma$. This subgroup is the kernel of the map to $\mathbb{Z}/n$ that kills $\Gamma$. – HJRW Apr 29 '12 at 19:24

The group $G$ maps onto the free product $C_p*C_p*C_p$ of three cyclic groups of order $p$ (just send each $b_i^p$ to $1$). This free product is virtually free, as a free product of finite groups (by Kurosh theorem the kernel of the homomorphism onto $C_p \times C_p \times C_p$ is free) and is not virtually cyclic. Hence it is large, and so $G$ is large as well. This works for any $p>1$.