G/Z(G) abelian and |Z(G)|=n odd ⇒ (xy)^n=x^ny^n for all x,y in G

Posted: March 22, 2022 in Basic Algebra, Groups
Tags: n-abelian groups

Here we defined an $n$ –abelian group as a group $G$ in which $(xy)^n=x^ny^n$ for all $x,y \in G.$ In this post, we see an interesting class of $n$ -abelian groups.

Problem. Let $G$ be a group (not necessarily finite) with the center $Z(G).$ Suppose that $G/Z(G)$ is abelian and $|Z(G)|=n.$

i) Show that if $n$ is odd, then $(xy)^n=x^ny^n$ for all $x,y \in G,$

ii) Show that i) is not necessarily true if $n$ is even.

Solution. Let $x,y \in G.$ First, let’s see what $G/Z(G)$ being abelian tells us about powers of $xy.$ Well, since $G/Z(G)$ is abelian, $xyZ(G)=yxZ(G)$ and so $xy=yxz$ for some $z \in Z(G).$ Thus

$yx=xyz^{-1}. \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Now suppose that

$(xy)^k=x^ky^kz^{a_k}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

for some integers $k \ge 1$ and $a_k.$ Note that $a_1=0.$ Now,

$(xy)^{k+1}=x(yx)^ky=x(xyz^{-1})^ky, \ \ \ \ \ \ \ \ \ \ \ \text{by} \ (1)$

$=x(xy)^kyz^{-k}, \ \ \ \ \ \ \ \ \ \ \text{because} \ z \in Z(G)$

$=x(x^ky^kz^{a_k})yz^{-k}, \ \ \ \ \ \ \ \ \ \ \ \ \text{by} \ (2)$

$=x^{k+1}y^{k+1}z^{a_k-k}=x^{k+1}y^{k+1}z^{a_{k+1}},$

and so the sequence $\{a_k\}_{k \ge 1}$ satisfies the conditions $a_1=0, \ a_k-k=a_{k+1}, \ k \ge 1.$ It is now easy to find a formula for $a_k$

$\displaystyle a_k=\sum_{j=1}^{k-1}(a_{j+1}-a_j)=-\sum_{j=1}^{k-1}j=\frac{k(1-k)}{2}.$

Hence, by $(2),$

$\forall x,y \in G, \ \exists z \in Z(G): \ \ \ \ (xy)^k=x^ky^kz^{\frac{k(1-k)}{2}}, \ \ \ \ \forall k \ge 1. \ \ \ \ \ \ \ \ \ (3)$

The problem is now easy to solve.

i) Let $x,y \in G.$ Since $n=|Z(G)|$ is odd, $n$ divides $\displaystyle \frac{n(1-n)}{2}$ and so $z^{\frac{n(1-n)}{2}}=1$ for all $z \in Z(G).$ Thus, since, by $(3),$ there exists $z \in Z(G)$ such that $(xy)^n=x^ny^nz^{\frac{n(1-n)}{2}},$ we have $(xy)^n=x^ny^n.$

ii) A counter-example is $G=D_8,$ the dihedral group of order $8.$ By this post, $|Z(G)|=2,$ which is an even number, and so $G/Z(G)$ is abelian because $|G/Z(G)|=4.$ But it is not true that $(xy)^2=x^2y^2$ for all $x,y \in G.$ For example, if we choose $x,y$ to be generators of $G,$ where $o(x)=2, \ o(y)=4, \ xyx=y^{-1},$ then $(xy)^2=1$ but $x^2y^2=y^2 \ne 1. \ \Box$

Example. Let $G$ be a group of order $p^3,$ where $p > 2$ is a prime. Show that $(xy)^p=x^py^p$ for all $x,y \in G.$

Solution. Nothing to prove if $G$ is abelian. Suppose now that $G$ is non-abelian. Since $G$ is a $p$ -group, we have $|Z(G)| > 1$ and so either $|Z(G)|=p$ or $|Z(G)|=p^2.$ Since $G$ is non-abelian, $G/Z(G)$ is not cyclic and so $|G/Z(G)| \ne p,$ which gives $|Z(G)| \ne p^2.$ So $|Z(G)|=p$ and hence $|G/Z(G)|=p^2,$ Thus $G/Z(G)$ is abelian and the result now follows from the first part of the above Problem. $\Box$