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A finite additive group is a finite group whose operation is written as addition. Its identity element is denoted $0$, the inverse element of $x$ is denoted $-x$, and translation $y\mapsto x+y$ is a bijection.

A group operation is the associative binary operation of a group. It has an identity element, and every element has an inverse element.

For a finite group $G$ and a subgroup $H$, Lagrange's theorem statesIn particular $|H|$ divides $|G|$: the left cosets partition $G$, and multiplication by a representative is a bijection from $H$ to each coset. Apply this to the cyclic subgroup $⟨g⟩$. Its size is the order of a group element $g$, so every element order divides $|G|$.

Under the square condition, every group element is its own inverse element. Consequentlyfor all $g,h$. This proves that a group of exponent two is abelian, including the trivial group.

For the fourth-power condition a counterexample to commutativity is the quaternion groupHere $i^2=j^2=k^2=-1$ and $ij=k=-ji$. Thus its six elements outside $\{\pm 1\}$ have order of a group element four, while $1$ and $-1$ have orders one and two. All fourth powers are the identity, yet $ij\ne ji$.

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:and both have first component $x+y+z$. The identity element is $(0,0)$, and the two-sided inverse element isThus 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 $a\ne 1$, 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. HenceBoth $H$ 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,so $K$ is a normal subgroup. For $H$, the corresponding calculation isWhen $a=1$ this lies in $H$. Conversely, for $a\ne 1$, take $x=0$, $u=1$, $y=1$: the second coordinate is $a-1\ne 0$, so $H$ is not normal. Thus $H$ is normal precisely when $a=1$.

Finally, the projectionis 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.

Left and right multiplication by a fixed group element are bijections, called left and right translations.