Please help me about the following question:
Suppose $G$ is a finite 2-group and $x\in Z(M)\setminus Z(G)$ for some maximal subgroup $M$ of $G$ such that $x^2\in Z(G)$, is it true $x\in Z_2(G)$? Where by $Z_2(G)$, I mean $Z(\frac{G}{Z(G)})$.
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$\begingroup$ I guess $Z_2(G)$ denotes the inverse image in $G$ of the center of $G/Z(G)$? $\endgroup$– YCorCommented Apr 4, 2018 at 15:00
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$\begingroup$ @YCor: Yes: it's standard notation for the upper central series. $Z_1(G) = Z(G)$, and $Z_{n+1}(G)$ is the inverse image of the center of $G/Z_n(G)$. $\endgroup$– Arturo MagidinCommented Apr 4, 2018 at 15:43
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1 Answer
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Yes. I hope this is not a homework exercise.
Sketch proof: $G = \langle M,t \rangle$ with $t^2 \in M$. Let $x^t = xa$; so $a \in Z(M)$. Then $x^2 = (x^2)^t = x^2a^2$, so $a^2=1$.
Also $x= x^{t^2} = xaa^t$, so $a^t=a$ and hence $a \in Z(G)$ and $x \in Z_2(G)$.
answered Apr 4, 2018 at 15:22
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$\begingroup$ many thanks. Your proof is very nice. $\endgroup$– MaryamCommented Apr 4, 2018 at 15:33