# Non abelian group of order 40 [closed]

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 $|Z(G)|=10$.

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

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• 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 – M Alrantisi Dec 14 '14 at 7:55

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 $\cong D_4 \times C_5$.
• 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 $D_4$. – Nicky Hekster Dec 14 '14 at 9:03
• $Z(Q)=\langle -1 \rangle$, where $Q=\{1,i,j,k,-1,-i,-,j,-k\}$. – Nicky Hekster Dec 14 '14 at 9:23