Prove that there exists a $z \in Z(G)$ such that $yx = zxy$.

Let $G$ be a non-abelian group of order 27. Assume $x, y \in G$.

Prove that there exists a $z \in Z(G)$ such that $yx = zxy$.

HINT: Start by determining $Z(G)$. Consider the cosets of $Z(G)$.

I managed to come up with a solution to this, but I did it without the hint provided.

What I did:

I noticed that the commutator subgroup $G'$ of $G$ is a subgroup of $Z(G)$ since the center is normal in $G$. Then there exists an element $z \in Z(G)$ such that $z = yxy^{-1}x^{-1}$ for some $x, y \in G$. Then $zxy = yx$ as desired.

My attempt with the hint: Since G is a $p$-group of order $27 = 3^{3}$ so the center has order 3. i.e., $Z(G) = \{1, z, z^{-1}\}.$

I don't know how to proceed from here.

Solution 1:

I noticed that the commutator subgroup G′ of G is a subgroup of Z(G) since the center is normal in G.

The statement is true, the justification is not. The commutator subgroup is not contained in every normal subgroup unless the group is Abelian.

With the hint. You already know that the center is of order $3$, so $G/Z(G)$ is of order $9$ and hence Abelian. Take $x,y\in G$. Then $x\in xZ(G), y\in yZ(G)$. The commutator of these two cosets must be then $Z(G)$ (the identity of $G/Z(G)$). So $[x,y]\in Z(G)$, so $xyx^{-1}y^{-1}=z\in Z(G)$ and you are done.

Solution 2:

Just note that since $G/Z(G)$ has order $9$, it must be abelian, so $\forall x, y \in G~(xyZ(G)=yxZ(G))$.