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I am trying to solve the following problem.

Let $G$ be a group. If $M, N \subset G$ are such that $x^{-1} M x = M$ and $x^{-1} N x = N$ for all $x \in G$ and $M \cap N = \{1\}$, prove that $m n = n m$ for all $m \in M, n \in N$.

I have already proven it for the specific case in which $M, N$ are subgroups of $G$ (see proof below). However, I can't prove it without this hypothesis. I think $M, N$ must be subgroups for the result to hold, but I can't find a counterexample to show this as well.

Any help with a proof or counterexample is much appreciated.

Current proof outline:

$$ (m \cdot n)\cdot (n \cdot m)^{-1} = (m \cdot n)\cdot (m^{-1} \cdot n^{-1}) =: k$$ $ k = (m \cdot n \cdot m^{-1}) \cdot n^{-1} \in N$, because $(m \cdot n \cdot m^{-1}) \in N $ and $n^{-1} \in N$ (and $N$ is subgroup).

$ k = m \cdot (n \cdot m^{-1} \cdot n^{-1}) \in M$, because $m \in M $ and $(n \cdot m^{-1} \cdot n^{-1}) \in M$ (and $M$ is subgroup).

So $k \in M \cap N = \{1\}$, which implies $k = 1$ and $$ (m \cdot n)\cdot (n \cdot m)^{-1} = 1 $$ $$ (m \cdot n)\cdot (n \cdot m)^{-1}\cdot (n \cdot m) = 1 \cdot (n \cdot m) $$ $$ m \cdot n = n \cdot m $$

Wheepy
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2 Answers2

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The assertion doesn't hold when $M$ or $N$ is not a subgroup of $G$. Here is a counterexample.

Let $M=\{e,(12),(13),(23)\}$ and $N=\{e, (123),(132)\}$ in $G=S_3$. Then $x^{-1}Mx=M$ and $x^{-1}Nx=N$ for all $x\in G$ and $M\cap N=\{e\}$. $M$ is not a subgroup of $G$ and $(123)(12)\neq (12)(123)$.

bfhaha
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Let $G$ be the dihedral group of order $8$, with $b$ an element of order $2$ and $a$ an element of order $4$. For $x,y\in G$ let $x^y=y^{-1}xy$. By direct calculation we have:

$$\begin{array}{rcc} x:&e&b&a&a^2&a^3&ab&a^2b&a^3b&\\ \hline a^x:&a&a^3&a&a&a&a^3&a^3&a^3\\ b^x:&b&b&a^2b&b&a^2b&a^2b&b&a^2b \end{array}$$

Let $M=\{1,a,a^3\}$ and $N=\{1,b,a^2b\}$; then $M$ and $N$ satisfy the stated conditions, but $ba=a^3b\ne ab$.

Brian M. Scott
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