## There does not exist a nilpotent group generated by two elements such that every nilpotent group generated by two elements is a quotient

Prove that there cannot be a nilpotent group $N$ generated by two elements such that every nilpotent group which is generated by two elements is a quotient of $N$. (That is, the specification of $c$ in the definition of free nilpotent groups is necessary.)

Suppose $N$ is such a group. Recall that $D_{2 \cdot 2^k}$ is generated by two elements and has nilpotence class $k$. By our hypothesis, there is a surjective group homomorphism $\varphi : N \rightarrow D_{2 \cdot 2^k}$. Now $\varphi[N^t] = \varphi[N]^t = D_{2 \cdot 2^k}^t$ for all $t$. Suppose $N$ has nilpotence class $c$; then $1 = \varphi[1] = \varphi[N^c]$ $= D_{2 \cdot 2^{c+1}}^c \neq 1$, a contradiction. Thus no such $N$ exists.