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

Question

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

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.

Just note that since $G/Z(G)$ has order $9$, it must be abelian, so $xyZ(G)=yxZ(G)$. — Robert Shore, Dec 16 '21 at 20:05
How do you conclude that the commutator subgroup is contained in the center? This is not true for all groups of order $27$. — Servaes, Dec 16 '21 at 20:07
@Servaes I read in a text that if a non-trivial subgroup is normal then the commutator subgroup is a subset of that subgroup. Since, the center of a group is normal in that group, I applied that here. — Malcolm, Dec 16 '21 at 20:18

score 3 · Accepted Answer · answered Dec 16 '21 at 20:18

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.

score 1 · Answer 2 · answered Dec 16 '21 at 20:11

1

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

answered Dec 16 '21 at 20:11

Robert Shore

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

2 Answers2