Intersection of a 2-generated subgroup with the commutator subgroup of the group

Posted: May 6, 2023 in Basic Algebra, Groups
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Let G be a group, and x,y \in G. Recall that the commutator subgroup G' of G is the subgroup generated by the set \{[a,b]: \ a,b \in G\}, where [a,b]:=aba^{-1}b^{-1}. So xyx^{-1}y^{-1} \in G'. Now, let’s consider the element z:=x^my^nx^py^q, where m,n,p,q are any integers, and ask: when do we have z \in G' ? Well, using the trivial identity ab=[a,b]ba, we have

z=x^my^nx^py^q=[x^m,y^n]y^nx^mx^py^q=[x^m,y^n]y^nx^{m+p}y^q=[x^m,y^n][y^n,x^{m+p}]x^{m+p}y^ny^q

=[x^m,y^n][y^n,x^{m+p}]x^{m+p}y^{n+q} \in G'x^{m+p}y^{n+q},

and so z \in G' if and only if x^{m+p}y^{n+q} \in G'. In particular, if o(x),o(y), the orders of x,y, are finite and o(x) \mid m+p, \ o(y) \mid n+q, then x^{m+p}y^{n+q}=1 \in G' and hence z \in G'. This can be easily extended to any element of the subgroup \langle x,y\rangle, using induction, to get the following.

Proposition. Let G be a group, and x,y \in G. Let z:=x^{m_1}y^{n_1} \cdots x^{m_k}y^{n_k}, where m_j,n_j, \ 1 \le j \le k, are any integers. Then

\displaystyle z=\left(\prod_{j=1}^{k-1} [x^{m_1+m_2+ \cdots +m_j},y^{n_1+ \cdots + n_j}][y^{n_1+ \cdots + n_j},x^{m_1+ \cdots + m_{j+1}}]\right)x^{m_1+ \cdots + m_k}y^{n_1+ \cdots + n_k},

and so z \in G' if and only if x^{m_1+ \cdots + m_k}y^{n_1+ \cdots + n_k} \in G'. In particular, if o(x),o(y) are finite and

\displaystyle o(x) \mid \sum_{j=1}^km_j, \ \ \ \ \ \ o(y) \mid \sum_{j=1}^k n_j,

then z \in G'. \ \Box

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