= Solution
Work with $x$ modulo $p-1$ and $u$ modulo $p$. By <Fermat's little theorem>, $a^{p-1}=1$ in $\mathbb F_p$, so the factor $a^y$ is independent of the chosen representative of $y$. The proposed multiplication is therefore well defined on the stated <Cartesian product>. Its two associative bracketings have the same second component:
$$
a^z(a^yu+v)+w=a^{y+z}u+a^zv+w,
$$
and both have first component $x+y+z$. The <identity element> is $(0,0)$, and the two-sided <inverse element> is
$$
\boxed{(x,u)^{-1}=(-x,-a^{-x}u)}.
$$
Thus the multiplication defines the <twisted cyclic pair group>, of order $p(p-1)$.
If $a=1$, multiplication is coordinatewise addition, so the <group> is abelian. If $a\ne1$, then $p>2$ and both $(1,0)$ and $(0,1)$ are available; their products in opposite orders are $(1,1)$ and $(1,a)$. They differ. Hence
$$
\boxed{G\text{ is abelian}\iff a=1.}
$$
Both $H$ and $K$ contain the <identity element> and are closed under the multiplication and inverses: they are the <cyclic groups> of orders $p-1$ and $p$ respectively. Conjugation gives, with all coordinates reduced in the appropriate modulus,
$$
(x,u)(0,v)(x,u)^{-1}=(0,a^{-x}v),
$$
so $K$ is a <normal subgroup>. For $H$, the corresponding calculation is
$$
(x,u)(y,0)(x,u)^{-1}=(y,a^{-x}(a^y-1)u).
$$
When $a=1$ this lies in $H$. Conversely, for $a\ne1$, take $x=0$, $u=1$, $y=1$: the second coordinate is $a-1\ne0$, so $H$ is not normal. Thus \b[$H$ is normal precisely when $a=1$].
Finally, the projection
$$
\boxed{\pi:G\longrightarrow\mathbb Z/(p-1)\mathbb Z,\qquad\pi(x,u)=x}
$$
is a surjective <group homomorphism> because the first coordinates add. Its <kernel of a group homomorphism> is exactly $K$. At $p=2$, $a=1$ is the only possibility and the first factor is trivial, consistent with every conclusion. No assumption that $a$ generates the multiplicative <group> of the <finite field> is needed.
Back to article page
