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I think it will in the lines we construct the dihedral groups. The group generated by $a,b$ such that $a^q=e$ , $b^p=e$ and one more condition.

Aritriya Mukhapadhayay
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    For the dihedral case, $p=2$ and you need the relation $aba^{-1}=b^{-1}$. The key here is that $-1$ is a nontrivial solution of $x^2\equiv1\pmod q$. – Angina Seng Jun 03 '20 at 02:36

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Take a nontrivial homomorphism $\varphi$ from $\Bbb Z_p$ to $\rm{Aut}(\Bbb Z_q)\cong\Bbb Z_q^\times\cong \Bbb Z_{q-1}$. Use it to construct the semidirect product $\Bbb Z_p{}_\varphi\ltimes\Bbb Z_q$.

This product will not be abelian. Consider: $(1,0)(0,1)=(1+0,0+\varphi_1(1))=(1,\varphi_1(1))$ but $(0,1)(1,0)=(0+1,1+\varphi_0(0))=(1,1)$.