= Solution
{c}
Right multiplication gives a right action $(Hx)\cdot g=Hxg$. To obtain a <homomorphism> into <permutations> with the usual function-composition convention, use the associated left action
$$
\rho(g)(Hx)=Hxg^{-1}.
$$
This is well-defined: replacing $x$ by $hx$, $h\in H$, does not change the image <coset>. Moreover $\rho(g)\rho(k)(Hx)=Hxk^{-1}g^{-1}=\rho(gk)(Hx)$, and each map is bijective with inverse $\rho(g^{-1})$.
An element $g$ is in the <kernel> precisely when $Hxg^{-1}=Hx$ for every $x\in G$. This says $xg^{-1}x^{-1}\in H$, equivalently $g\in x^{-1}Hx$. Hence
$$
\boxed{\ker\rho=\bigcap_{x\in G}x^{-1}Hx.}
$$
This is the largest <normal subgroup> of $G$ contained in $H$.
Now the index of $H$ is $q$, so $G/\ker\rho$ embeds into $S_q$. Its order therefore divides $q!$. Since $p>q$ is <prime>, $p$ does not divide $q!$. From $|G|=pq$ it follows that $p$ divides $|\ker\rho|$. But $\ker\rho\subseteq H$ and $|H|=p$, forcing $\ker\rho=H$. Therefore \b[$H$ is normal].
For the last case let $H=\langle b\rangle$. Normality makes <conjugation> by $a$ an <automorphism>, so $a^{-1}ba=b^r$ with $r\in\{1,2,3,4\}$. Applying it twice and using $a^2=1$ gives $b=b^{r^2}$, so $r^2\equiv1\pmod5$. Thus $(r-1)(r+1)\equiv0\pmod5$, and
$$
\boxed{a^{-1}ba=b\ \text{or}\ b^{-1}.}
$$
Both possibilities are allowed; the order assumptions alone do not force the inverse case.
Back to article page
