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 Idrissi Dec 29 '14 at 8:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hanul Jeon, user26857, DeepSea, Hakim, Najib Idrissi
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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

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 $\cong D_4 \times C_5$.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.