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Let $G$ be a non abelian group of order 40 that is the direct product of two of its Sylow subgroups. Show that the center of $G$ has order 10.

I tried 40=5×8 Since $G$ not abelian $G$ NOT EQUAL CENTER OF $G$. Then stuck how to interfere Sylow theorem to find $|Z(G)|=10 $.

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closed as off-topic by Hanul Jeon, user26857, DeepSea, Hakim, Najib Idrissi Dec 29 '14 at 8:16

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  • $\begingroup$ I tried 40=5×8 Since G not abelian G NOT EQUAL CENTER OF G .then stuck how to interfere sylow theorem to find |Z(G)|=10 $\endgroup$ – M Alrantisi Dec 14 '14 at 7:55
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Hint: there are only two non-abelian groups of order $8$, the quaternion group or the dihedral. So $G \cong Q \times C_5$ or $\cong D_4 \times C_5$.

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  • $\begingroup$ And how I get 10 been struggling to show it can you please explain it $\endgroup$ – M Alrantisi Dec 14 '14 at 8:50
  • $\begingroup$ In general $Z(G_1 \times G_2) = Z(G_1) \times Z(G_2)$. So you are left with calculating the centers of $Q$ and $D_4$. $\endgroup$ – Nicky Hekster Dec 14 '14 at 9:03
  • $\begingroup$ Centre of D4 is r^2 buthe I don't know how to find Q group centre I read that Q 8 centre are 1,-1 $\endgroup$ – M Alrantisi Dec 14 '14 at 9:18
  • $\begingroup$ $Z(Q)=\langle -1 \rangle$, where $Q=\{1,i,j,k,-1,-i,-,j,-k\}$. $\endgroup$ – Nicky Hekster Dec 14 '14 at 9:23
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    $\begingroup$ Of course you don't really need a classification of groups of order 8, it suffices to note that the center can only have order 2 or 8 because of the classical observations that the center must be non-trivial in a p-group and if G/Z(G) is cyclic, it is actually trivial. $\endgroup$ – Myself Dec 14 '14 at 10:14

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