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I am trying to solve the following problem:

Let $G$ be a finite group such that there exists an isomorphism $\phi:f:G/Z(G) \to \mathbb Z_3 \oplus \mathbb Z_3$ and $x \in G$ such that $f(\overline{x})=(1,0)$. Show that for $y \in G$, $$xy=yx \leftrightarrow f(\overline{y}) \in \langle (1,0) \rangle$$

I could prove the left implication, if $f(\overline{y}) \in \langle (1,0) \rangle$, then $f(\overline{y})=kf(\overline{x})$ for $k=0,1,2$. From here we have the cases $y,yx^{-1},yx^{-2} \in Z(G)$ and from here I could prove that $yxy^{-1}x^{-1}=e$.

I don't know what to do to show the other implication. I know $f(\overline{y})=k(1,0)+t(0,1)$ since $\mathbb Z_3 \oplus \mathbb Z_3=\langle (1,0),(0,1) \rangle$ but I don't know how to deduce that $t=0$. Any hints would be greatly appreciated.

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  • $\begingroup$ The important point is that the quotient is only by the center, and no more or less. For example, if $G = \mathbb Z_3 \oplus \mathbb Z_3$, then there is a map $f: G \to \mathbb Z_3 \oplus \mathbb Z_3$ whose kernel is central, but it does not present the latter as the quotient of the former by the center. $\endgroup$ – Theo Johnson-Freyd Jul 3 '15 at 17:39
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    $\begingroup$ What this means in practice is the following. Suppose $y$ commutes with $x$, and $\bar x = (1,0)$. Suppose also that $\bar y = (*,1)$. Then $y$ commutes with $x,y,Z(G)$ (because everything commutes with $Z(G)$). Now prove that $\langle x,y,Z(G)\rangle = G$. So $y$ is central. But $\bar y \neq 0$, so $y$ is not central. $\endgroup$ – Theo Johnson-Freyd Jul 3 '15 at 17:45
  • $\begingroup$ Thanks for your help! Take $t \in G$ and suppose $f(\overline{t})=(a,b)$. We can write $(a,b)=(a-b*)(1,0)+b(*,1)=(a-b*)f(\overline{x})+bf(\overline{y})=f(\overline{x^{a-b*}y^{b}})$. Since $f$ is injective, we have $\overline{t}=\overline{x^{a-b*}y^{b}}$, in other words, $ty^{-b}x^{b*-a}=z$, with $z \in Z(G)$. So $t=zx^{a-b*}y^{b} \in \langle Z(G),x,y \rangle$. From here it is easy to prove that $y$ is central, which means $f(\overline{y})=(0,0)$, a contradiction. $\endgroup$ – user16924 Jul 3 '15 at 19:40
  • $\begingroup$ Btw, I couldn't understandyour first comment: "...whose kernel is central, but it does not present the latter as the quotient of the former by the center.", I would like to grasp this and not just the mechanics of the exercise, could you explain this to me again? $\endgroup$ – user16924 Jul 3 '15 at 19:47
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    $\begingroup$ I meant only that, not knowing exactly how you were thinking about the problem, I assumed that you were using the fact that $f : G \to \mathbb Z_3 \oplus \mathbb Z_3$ was a homomorphism (here by "$f$" I mean the pullback of your $f$ along the canonical map $G \to G/Z(G)$), and that $\ker f \subseteq Z(G)$, but that you had not yet used that $\ker f = Z(G)$. My comment was only that you need to use that $\ker f = Z(G)$ precisely, and not just $\subseteq$ or $\supseteq$. $\endgroup$ – Theo Johnson-Freyd Jul 4 '15 at 17:52

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