Daily math problems for May 18, 2026 — Algebra, Elementary Functions, Vectors

Take any $g\mathrm{\ne }e$. The subgroup $⟨g⟩$ has order dividing $p$. It's not $1$ (since $g\mathrm{\ne }e$), so it has order $p$, i.e. $⟨g⟩=G$. So $G$ is cyclic, hence abelian.

$x^2>9\iff \mathrm{\mid }x\mathrm{\mid }>3\iff x<-3$ or $x>3$. Solution: $(-\mathrm{\infty },-3)\cup (3,+\mathrm{\infty })$.

$3\cdot (-5)=(m-1)(-(m+1))$, i.e. $-15=-(m-1)(m+1)=1-m^2$, so $m^2=16$ and $m\in \{-4,4\}$.