2. A preliminary lemma

We need the following lemma in the proof of our main result.

Lemma 2·1 Suppose that $x_1,x_2,x_3,g_1,g_2,g_3$ are elements of a group such that $[x_1u_1,x_2u_2,x_3u_3]=1$ for each triple of the following triples $(u_1,u_2,u_3)$ :

\begin{eqnarray*} &&(1,1,1),(g_1,g_2,g_3),(g_1,g_2,1),(g_1,1,g_2);\\[2pt] &&(g_1,1,1),(g_1,1,g_3),(1,1,g_1),(1,g_2,g_1);\\[2pt] && (1,g_2,1),(1,1,g_2),(1,g_2,g_3),(1,1,g_3).\end{eqnarray*}

Then $[g_1,g_2,g_3]=1$ .

Proof. Note that [x,y] denotes $x^{-1}y^{-1}xy$ and $[x,y,z]:=[[x,y],z]$ for arbitrary elements x,y,z of a group. We will throughout use famous commutator calculus identities: $[xy,z]=[x,z]^y[y,z]$ ( $\dagger$ ) and $[x,yz]=[x,z][x,y]^z$ ( $\dagger \dagger$ ) for all elements x,y,z of a group, where $g^h$ denotes $h^{-1}gh$ . In the following (i) refers to the equality $[x_1u_1,x_2u_2,x_3u_3]=1$ , where $(u_1,u_2,u_3)$ is the ith triple counting them from left to right starting at the top.

\begin{eqnarray*} 1&=&[x_1g_1,x_2g_2,g_3]=[[x_1g_1,g_2][x_1g_1,x_2]^{g_2},g_3] \;\; {\textrm{by}} \; (\dagger \dagger), \; (2) \; {\textrm{and}} \; (3) \\[2pt] &=&[[x_1g_1,g_2][x_1g_1,x_2],g_3] \;\; {\textrm{by}} \; (4) \; {\textrm{and}} \; (5) \\[2pt] &=&[x_1g_1,g_2,g_3]=[[x_1,g_2]^{g_1}[g_1,g_2],g_3] \;\; {\textrm{by}} \; (\dagger), \; (5) \; {\textrm{and}} \; (6). \;\;\;\; {\textrm{(I)}} \end{eqnarray*}

On the other hand,

\begin{eqnarray*} 1&=&[x_1,x_2g_2,g_1] \;\; {\textrm{by}} \; (8) \; {\textrm{and}} \; (9)\\[2pt] &=&[[x_1,g_2][x_1,x_2]^{g_2},g_1]=[[x_1,g_2][x_1,x_2],g_1] \;\; {\textrm{by}} \; (\dagger \dagger), \; (1) \; {\textrm{and}} \; (10)\\[2pt] &=&[x_1,g_2,g_1] \;\; {\textrm{by}} \; (1) \; {\textrm{and}} \; (7). \;\;\;\; {\textrm{(II)}} \end{eqnarray*}

Also,

\begin{eqnarray*} 1&=&[x_1,x_2g_2,g_3] \;\; {\textrm{by}} \; (9) \; {\textrm{and}} \; (11)\\[2pt] &=&[[x_1,g_2][x_1,x_2]^{g_2},g_3]=[[x_1,g_2][x_1,x_2],g_3] \;\; {\textrm{by}} \; (\dagger \dagger), \; (1) \; {\textrm{and}} \; (10)\\[2pt] &=&[x_1,g_2,g_3] \;\; {\textrm{by}} \; (1) \; {\textrm{and}} \; (12). \;\;\;\; {\textrm{(III)}}\end{eqnarray*}

Now it follows from (I), (II) and (III) that $[g_1,g_2,g_3]=1$ .

Remark. The “left version” ( $g_ix_j$ instead of $x_jg_i$ ) of Lemma 1·2 is not clear to hold. The validity of a similar result to Lemma 1·2 for commutators with length more than 3 is also under question.

3. Compact groups with many elements satisfying the 2-step nilpotent law

We need the “right version” of [Reference Soleimani Malekan, Abdollahi and Ebrahimi5, theorem 2·3] as follows.

Proof. Since $\mathbf{m}_G(A)=\mathbf{m}_G(A^{-1})$ , it follows from [Reference Soleimani Malekan, Abdollahi and Ebrahimi5, theorem 2·3] that there exists an open subset V of G containing 1 such that

\[\mathbf{m}_G(A^{-1} \cap v_1A^{-1} \cap \cdots \cap v_kA^{-1})>0\]

for all $v_1,\dots,v_k \in V$ . Now take $U:=V^{-1}$ which is an open subset of G containing 1. Thus for all $u_1,\dots,u_k \in U$

\begin{eqnarray*} \mathbf{m}_G(A \cap Au_1 \cap \cdots \cap Au_k) &=&\mathbf{m}_G((A \cap Au_1 \cap \cdots \cap Au_k)^{-1})\\[2pt] &=&\mathbf{m}_G(A^{-1} \cap u_1^{-1}A^{-1} \cap \cdots \cap u_k^{-1}A^{-1})>0 \end{eqnarray*}

This completes the proof.

Now we can prove our main result.

Proof of Theorem 1·2. Let $X:={\mathcal N}_2(G)$ . It follows from Theorem 3·1 and [Reference Hewitt and Ross2, theorem 4·5] that there exists an open subset $U=U^{-1}$ of G containing 1 such that

(*) \begin{align}X \cap X\bar{u}_1 \cap \cdots \cap X\bar{u}_{11}\neq \varnothing \end{align}

for all $\bar{u}_1,\dots,\bar{u}_{11} \in U\times U \times U$ . Now take arbitrary elements $g_1,g_2,g_3 \in U$ and consider

\begin{eqnarray*} \bar{u}_1&=&(g_1^{-1},g_2^{-1},g_3^{-1}), \bar{u}_2=(g_1^{-1},g_2^{-1},1), \bar{u}_3=(g_1^{-1},1,g_2^{-1})\\[2pt] \bar{u}_4 &=&(g_1^{-1},1,1),\bar{u}_5=(g_1^{-1},1,g_3^{-1}), \bar{u}_6=(1,1,g_1^{-1}),\bar{u}_7=(1,g_2^{-1},g_1^{-1})\\[2pt] \bar{u}_8 &=&(1,g_2^{-1},1), \bar{u}_9=(1,1,g_2^{-1}), \bar{u}_{10}=(1,g_2^{-1},g_3^{-1}), \bar{u}_{11}=(1,1,g_3^{-1}). \end{eqnarray*}

By $(*)$ , there exists $(x_1,x_2,x_3)\in X$ such that all the following 3-tuples are in X:

\begin{eqnarray*} &&(x_1g_1,x_2g_2,x_3g_3),(x_1g_1,x_2g_2,x_3),(x_1g_1,x_2,x_3g_2) \\[2pt] &&(x_1g_1,x_2,x_3),(x_1g_1,x_2,x_3g_3),(x_1,x_2,x_3g_1),(x_1,x_2g_2,x_3g_1)\\[2pt] &&(x_1,x_2g_2,x_3),(x_1,x_2,x_3g_2),(x_1,x_2g_2,x_3g_3),(x_1,x_2,x_3g_3). \end{eqnarray*}

Now Lemma 2·1 implies that $[g_1,g_2,g_3]=1$ . Therefore the subgroup $H:=\langle U \rangle$ generated by U is 2-step nilpotent. Since $H=\bigcup _{n\in {\mathbb N}} U^n$ , H is open in G. This completes the proof.

We finish with the following open question that would resolve Question 1·1 for arbitrary k: