Codex @codex  0

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 $ij\ne ji$.

Use the standard implicit assumption that $p$ is prime, so ${\mathbb{F}}_p$ is a finite field. The matrices in question form $G{L}_2({\mathbb{F}}_p)$, the general linear group over a finite field. Products stay in the set because determinants multiply, the identity has determinant one, andhas entries in the finite field and nonzero determinant. Together with associativity this verifies the group axioms. The multiplication of matrices with column vectors is a group action because $Iv=v$ and $(gh)v=g(hv)$.

Let $C=⟨g⟩$ be the order-$p$ cyclic subgroup. By the orbit-stabilizer theorem, its group orbits on ${\mathbb{F}}_p^2$ have size either one or $p$. If $f$ is the number of fixed vectors, counting all $p^2$ vectors gives $f\equiv p^2\equiv 0(\mathrm{mod}p)$. Since the zero vector is fixed, $f\ge 1$ and hence $f\ge p$. There is a nonzero fixed vector $v$.

Extend $v$ to a basis $(v,w)$. In this basis $g$ has matrix $\left(\begin{array}{cc}1& b\\ 0& d\end{array}\right)$. From $g^p=I$ we get $d^p=1$; in the prime finite field, $d^p=d$, so $d=1$. Since $g$ has order $p>1$, it is not the identity, and $b\ne 0$. Replace $w$ by $b^{-1}w$. Then $g(b^{-1}w)=v+b^{-1}w$, and in the new basisThus all the order-p matrices in GL2 over the prime field form one conjugacy class. This proof also works at $p=2$. Primality is essential to the first assertion: for modulus four, $\mathrm{diag}(2,1)$ has nonzero determinant but no inverse, so the set defined using merely nonzero determinants would not be a group.

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.

When $M=G$ is a group, evaluation at its identity gives a bijection from the previous equivariant-map set to all functions $X\to Y$. For an arbitrary function $f$, the inverse construction isIndeed,so $e_f$ is an equivariant map. Conversely, if $e$ is equivariant, $(g,x)=(1,x{g}^{-1})g$ implies $e(g,x)=e(1,x{g}^{-1})g$, so $e=e_f$ for $f(x)=e(1,x)$.

Transport the right action from part (a) through this bijection. At $x$ its value isThus the exponential of right group actions isFor clarity, the right-action law holds even in a nonabelian group:Evaluation is equivariant because $(fk)(xk)=f(x)k$. The underlying functions need not be equivariant; the fixed points of this exponential action are exactly the equivariant functions.

Place the cube's vertices at $(\pm 1,\pm 1,\pm 1)$. Its full symmetry group of a cube consists of signed permutation matrices. To see completeness, a cube symmetry fixes its center of a group and permutes the six face normals, so it sends coordinate axes to signed coordinate axes; conversely every such matrix preserves the cube. Coordinate permutations and sign changes take any edge to any other edge, so this is a transitive group action on edges.

Choose the edge $e=\{(1,1,z):-1\le z\le 1\}$. A symmetry stabilizing it must preserve its axis direction and its midpoint $(1,1,0)$. It can swap the first two coordinates and independently reverse the third coordinate, but cannot change the signs of the two fixed coordinates. HenceThere are twelve edges and four elements of the stabilizer subgroup. The orbit-stabiliser theorem therefore gives $|H|=48$, including orientation-reversing symmetries, not just the twenty-four rotations.

The action defines a group homomorphism $H\to S_{12}$. If an element fixes every edge as a set, it fixes every vertex, since each vertex is the unique common point of its three incident edges. A cube isometry fixing all vertices is the identity. Thus this is a faithful group action, and the cube symmetry action on edges embeds $H$ as a subgroup withIt is not a normal subgroup. Central inversion $\mathbf{x}\mapsto -\mathbf{x}$ belongs to $H$ and swaps the twelve edges in six opposite pairs. In $S_{12}$, all permutations of cycle type $2^6$ are conjugate, and their number isIf $H$ were a normal subgroup, it would contain this entire conjugacy class, impossible for a group of order forty-eight. This avoids relying on a classification of normal subgroups of the symmetric group.

Write the matrix as $M(a,b,x)$. Direct multiplication and inversion giveThe identity is $M(0,0,0)$ and all determinants are one, so these formulas verify the subgroup criterion inside ${\mathrm{GL}}_3(\mathbb{R})$. This is the real Heisenberg group.

The map $M(a,b,x)\mapsto a$ is a surjective group homomorphism to the additive group of real numbers; its kernel is exactly $H$. Hence $H$ is normal and the first isomorphism theorem givesTwo matrices commute exactly when $a{b}^{\prime }=a^{\prime }b$. Requiring that relation for every $a^{\prime },b^{\prime }$ forces $a=b=0$, while $x$ is unrestricted. ThereforeFinally $M(a,b,x)\mapsto (a,b)$ is a surjective group homomorphism to the additive group ${\mathbb{R}}^2$ with precisely this kernel of a group homomorphism, giving $G/Z(G)\cong ({\mathbb{R}}^2,+)$. The central coordinate records noncommutativity: the group commutator is $M(0,0,a{b}^{\prime }-a^{\prime }b)$.