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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.

A real Lie group is a finite-dimensional real smooth manifold with a group structure whose multiplication and inversion are smooth. Its tangent space at the identity element $e$ is the vector space $\mathfrak{g}=T_{e}G$ of velocities of smooth curves through $e$. For the general linear group ${\mathrm{GL}}_n(\mathbb{R})$ it identifies with $M_n(\mathbb{R})$. The matrix exponential and local matrix logarithm areThe exponential is globally defined, and each matrix $\exp X$ is invertible with inverse $\exp (-X)$. The logarithm here is a local map near $I$; it is not a globally defined inverse on every real invertible matrix. The differential of $\exp$ at zero is the identity, and the two power series are inverse locally.

For a general $G$, the assumed Exponential map of a Lie group with $d{\exp }_0=I$ has a smooth local inverse by the inverse function theorem, giving a local exponential chart. Define the Lie bracket from local group commutators byFor small $s,t$ the group commutator lies in that chart. More explicitly, the map $F(u,v)=\log (\exp u\exp v\exp (-u)\exp (-v))$ is smooth and has $F(u,0)=F(0,v)=0$. Its mixed second differential at $(0,0)$ is a bilinear map of $X,Y$, proving bilinearity. Swapping the two group elements inverts the commutator, and $\log (g^{-1})=-\log g$ near $e$. Thus $F(v,u)=-F(u,v)$ and $[Y,X]=-[X,Y]$. In the matrix group, expansion to the mixed term $st$ gives the familiar commutator $XY-YX$.

For $g\in G$, let $c_g(h)=gh{g}^{-1}$ and define the Adjoint representation of a Lie group by ${\mathrm{Ad}}_g=d(c_g{)}_e\in \mathrm{GL}(\mathfrak{g})$. It is smooth and satisfies ${\mathrm{Ad}}_{gh}={\mathrm{Ad}}_g{\mathrm{Ad}}_h$. Its derived representation is $\mathrm{ad}=d(\mathrm{Ad}{)}_e:\mathfrak{g}\to \mathrm{End}(\mathfrak{g})$. Naturality of the Exponential map of a Lie group under conjugation givesAt $t=0$, the derivative of the logarithm of the commutator is ${\mathrm{Ad}}_{\exp (sX)}Y-Y$: the derivative of multiplication at $(e,e)$ adds tangent vectors, and $d{\log }_e=I$. Differentiate in $s$ to concludeTo prove the Jacobi identity without assuming it in the bracket construction, first establish that the differential of a Lie group homomorphism preserves Lie brackets. For a homomorphism $f:G\to H$, the allowed exponential identity gives ${\log }_H(f({\exp }_{G}u))=d{f}_e(u)$ locally. Apply this identity to the group commutator and take the mixed derivative to obtain $d{f}_e([X,Y])=[d{f}_{e}X,d{f}_{e}Y]$. In particular $f=\mathrm{Ad}:G\to \mathrm{GL}(\mathfrak{g})$ givesApplying both sides to $Z$ yields $[[X,Y],Z]=[X,[Y,Z]]-[Y,[X,Z]]$. Rearranging with antisymmetry provesNeither injectivity of $\mathrm{ad}$ nor the existence of a global logarithm was used.

Finally let $H$ be a normal Lie subgroup of $G$, with its corresponding Lie algebra $\mathfrak{h}\subseteq \mathfrak{g}$. For each $g\in G$, conjugation restricts to $H$, so its differential preserves $\mathfrak{h}$: ${\mathrm{Ad}}_g\mathfrak{h}=\mathfrak{h}$. Differentiate along $g=\exp (tX)$ to get $[X,Y]=\mathrm{ad}(X)Y\in \mathfrak{h}$ for every $X\in \mathfrak{g}$ and $Y\in \mathfrak{h}$. Thus the Lie algebra of a normal Lie subgroup is an ideal of a Lie algebra. The assertion concerns a subgroup carrying the corresponding Lie-subgroup structure, for example any closed subgroup; no arbitrary abstract subgroup is being assigned a tangent space.

Suppose every group commutator belongs to $H$. For $h\in H$ and $g\in G$, the particular group commutator $h^{-1}{g}^{-1}hg$ belongs to $H$. Multiplying on the left by $h$ gives $g^{-1}hg\in H$. Thus $g^{-1}Hg\subseteq H$ for every $g$; applying the same inclusion to $g^{-1}$ gives the reverse inclusion. Therefore $H$ is a normal subgroup.

We can now form the quotient group. Its group commutators areA group commutator is the identity element exactly when the two elements commute. Hence $G/H$ is an abelian group. Equivalently, the hypothesis says that $H$ contains the commutator subgroup, so the quotient factors through the abelianization.

Use the class equation for the action of $G$ on itself by conjugation:where the sum runs over noncentral conjugacy classes. By Lagrange's theorem, each noncentral class has size a positive power of $p$ greater than one. Thus $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. Since the identity element belongs to the center of a group,For $|G|=p^3$, the possible orders of the center of a group are $p,p^2,p^3$. Order $p^3$ would make $G$ an abelian group. Order $p^2$ would make the quotient group $G/Z(G)$ have prime order and hence be a cyclic group. But a cyclic quotient by the center forces $G$ to be Abelian: if $gZ(G)$ generates the quotient, any two elements are $g^{a}z$ and $g^{b}w$ with $z,w\in Z(G)$, and their products commute. Both larger orders contradict the hypothesis. Consequently

The Noncommutative Ruzsa triangle inequality says that nonempty finite subsets $B,C,D$ of a group satisfyThe Ruzsa covering lemma says that if $|CD|\le K|D|$, then some $X\subseteq C$ with $|X|\le K$ satisfies

Now let $A$ be a symmetric subset of a group, so $A^{-1}=A$ and $1\in A$. Apply the triangle inequality with the middle set $A$ to obtain, for $m\ge 4$,The hypothesis therefore gives $|A^m|\le K|A^{m-1}|$. Starting from $|A^3|\le K|A|$ proves

In particular $|A^5|\le K^3|A|$. Apply the covering lemma to $C=A^4$ and $D=A$. There is $X\subseteq A^4$ with $|X|\le K^3$ such thatThe set $A^2$ is symmetric and contains the identity element, so this inclusion is exactly the covering condition showing that

Small doubling alone is insufficient in a noncommutative group. Let $H$ be a finite group, let $G=H\ast ⟨x⟩$ be its free product with an infinite cyclic group, and putThen $A$ is symmetric and $|A^2|\le 5|H|+4=O(|A|)$, while $A^3$ contains the double coset $HxH$. Distinct pairs $(h_1,h_2)\in H^2$ give distinct reduced words $h_{1}x{h}_2$, so $|HxH|=|H{|}^2$. Letting $|H|\to \infty$ proves the small doubling does not control tripling in a noncommutative group phenomenon.

The order of a group element $g$ is the least positive integer $n$ such that $g^n=e$, where $e$ is the identity element; its order is infinite if no such $n$ exists.

Let $g$ have finite order $n$. A group homomorphism preserves the group operation and the identity, soThe order of an element divides every positive exponent that gives the identity. Hence

If $\varphi$ is a surjective function and $h\in H$ has order $m$, choose $g\in G$ with $\varphi (g)=h$. The first result gives $m\mid n$, where $n=\mathrm{ord}(g)$. The elementthen has order $m$, because the order of $g^k$ is $n/\gcd (n,k)$.

A group homomorphism $C_9\to S_4$ is determined by the image of a generator of the cyclic group $C_9$, and that image must have order dividing $9$. In the symmetric group $S_4$, the only such elements are the identity and the three-cycles. There arethree-cycles. Therefore the number of homomorphisms is

A subset $A$ of a multiplicatively written group is symmetric when it contains the identity element and $A^{-1}=A$. Consequently $⟨A⟩={\bigcup }_{m\ge 1}{A}^m$.

A torsion-free group is a group in which the identity element is the only element of finite order of a group element.