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For a nontrivial finite p-group $G$ of order $p^m$ with $m\ge 1$, partition into conjugacy classes. A noncentral class has size $[G:C_G(x)]$, a positive power of $p$ larger than one by Lagrange's theorem. The class equation therefore saysThe center of a group contains the identity, so $|Z(G)|\ge p$, proving its nontriviality. The positive-exponent qualification matters: the one-element group has no nonidentity central element.

If $|G|=p^2$, its center of a group has order $p$ or $p^2$. In the first case, the quotient group $G/Z(G)$ has prime order and is cyclic. Whenever a central quotient group is cyclic, writing all elements as $g^{i}z$ with $z$ central shows that they commute: $(g^{i}z)(g^{j}w)=g^{i+j}zw=(g^{j}w)(g^{i}z)$. Thus that case would already make $G$ Abelian and its center of a group all of $G$, a contradiction. Hence every group of order $p^2$ is Abelian.

If there is an element of order $p^2$, it is a generator of a group for $G$, giving $C_{p^2}$. Otherwise every nonidentity element has order $p$. Pick $a\ne 1$ and $b\notin ⟨a⟩$. Their cyclic subgroups have trivial intersection, and they commute; the $p^2$ distinct products $a^i{b}^j$ exhaust $G$. Hence the classification of groups of order p squared isBoth groups exist and are nonisomorphic, since only the first has an element of order $p^2$.

Write $U^{\ast }={\overline{U}}^T$ for the conjugate transpose. Taking determinants in $U^{\ast }AU=A$ gives $|\det U{|}^2\det A=\det A$. Since the Hermitian matrix $A$ is invertible, every such $U$ is invertible and $|\det U|=1$. The identity belongs to $U_A$; if $U,V\in U_A$, then $(UV{)}^{\ast }A(UV)=V^{\ast }AV=A$; and multiplying $U^{\ast }AU=A$ by $(U^{\ast }{)}^{-1}$ and $U^{-1}$ gives $(U^{-1}{)}^{\ast }A{U}^{-1}=A$. Associativity comes from matrix multiplication, so this is a group.

The Hermitian form $h_A(v,w)=v^{\ast }Aw$ obeys $h_A(Uv,Uw)=h_A(v,w)$ precisely when $U\in U_A$. Thus $U_A$ is the group of linear isometries of this nondegenerate Hermitian form; positivity of $A$ is unnecessary.

For $A=I$, $U$ is a unitary matrix. Here is a direct diagonalization of a matrix proof. Over $\mathbb{C}$, its characteristic polynomial has a root, giving a unit eigenvector $v$ with $Uv=\lambda v$. Preservation of the inner product gives $|\lambda |=1$. If $w\perp v$, then $⟨Uw,Uv⟩=⟨w,v⟩=0$, so $v^{\perp }$ is invariant. The restriction is again a unitary matrix. Induction on the dimension supplies an orthonormal basis of eigenvectors. In that basis, $U$ is diagonal, with diagonal entries on the unit circle.

Let $R$ be rotation through $2\pi /3$ about $z$, and let $S$ be rotation through $\pi$ about $x$. These generate the order-six dihedral group with relationsnamely $\{I,R,R^2,S,RS,R^{2}S\}$. We specify its order explicitly because dihedral notation has two conventions. It is a proper finite subgroup of the full axial rotation group.

To see that it is sufficient, write a general matrix in blocks $T=\left(\begin{array}{cc}B& u\\ v^T& c\end{array}\right)$. Commutation with $R$ forces $u=v=0$, since a nontrivial planar $120$-degree rotation fixes no nonzero planar vector. Its planar part is $\cos (2\pi /3)I+\sin (2\pi /3)J$, with nonzero sine, so $B$ must commute with $J$ and has the form $aI+bJ$. Commutation with $S$ again kills $b$. Hence this six-element subgroup has precisely the required invariant tensors.

It is also smallest by order. Groups of orders $2,3,5$ are cyclic: a nonidentity element has order dividing the prime group order by Lagrange theorem. For a cyclic rotation subgroup, the nonzero antisymmetric tensor representing cross product with its rotation-axis direction is invariant, so it cannot force the symmetric form above. The trivial subgroup plainly imposes no constraint. A group of order four is either cyclic or has three nonidentity elements of order two; in the latter case they commute, since $(AB{)}^{-1}=AB$ and also equals $BA$. Commuting distinct half-turns in three dimensions have perpendicular axes: conjugation by one half-turn must preserve the other's axis, and a distinct preserved axis lies in its perpendicular plane. In the specified axial group, their three axes are consequently $z$ and two perpendicular horizontal directions. In coordinates along them every diagonal matrix is invariant, including one with unequal horizontal entries. Such a group again does not force the required form.

Thus the smallest subgroup has six elements, generated by the axial $120$-degree rotation and one horizontal half-turn. It is the smallest axial rotation group forcing second-rank transverse isotropy; conjugating the horizontal axis gives equally valid choices.

For a finite group, the conjugacy classes partition the group, and a class is a singleton exactly when its element belongs to the centre of a group. Since the centre is trivial, the class equation becomeswhere the ${\mathcal{C}}_j$ are all the nonidentity conjugacy classes. If the prime number $p$ divided every $|{\mathcal{C}}_j|$, reduction modulo $p$ would give $0=1$, because $p\mid |G|$. Therefore at least one of these class sizes is not divisible by $p$. Primality now givesIts size is greater than one because the centre is trivial. This proves the prime-to-p conjugacy class lemma.

The conclusion requires $p$ to be prime. The printed question does not explicitly impose this hypothesis. If arbitrary composite divisors are allowed, take the symmetric group $S_3$ and $p=6$: its centre is trivial, its two nonidentity conjugacy classes have sizes $3$ and $2$, and neither is coprime to $6$. Thus the argument proves the intended prime case and also identifies why the unrestricted literal reading is false.

We first prove the needed instance of Cauchy theorem for groups, rather than invoke it. Let a prime number $q$ divide $|G|$ and consider the setThere are $|G{|}^{q-1}$ such tuples: choose the first $q-1$ entries freely, and the last is forced. Cyclic rotation preserves this set, because moving the first entry to the end conjugates the product, which remains $1$. Each rotation orbit of a group action has either one or $q$ elements. Indeed a nontrivial shift fixing a tuple generates all shifts, since its step has an inverse modulo the prime $q$. The fixed tuples are precisely $(g,\dots ,g)$ with $g^q=1$.

Counting nonfixed orbits in multiples of $q$, and using $q\mid |G{|}^{q-1}$, shows that the number of solutions of $g^q=1$ is divisible by $q$. The identity supplies one solution, so there is a nonidentity solution. If its order of a group element is $d$, divide $q$ by $d$: the remainder would give a smaller positive exponent producing the identity, so $d\mid q$. Hence $d=q$. This proves the required prime-order existence result by the cyclic-tuple proof of Cauchy theorem.

Our hypothesis says that this nonidentity element has order $p$, so $q=p$. Every prime divisor of $|G|$ is therefore $p$, which givesThus a finite group of prime exponent has prime-power order. The case $n=0$ includes the trivial group. No group-theoretic counting theorem was assumed in the tuple argument.

A subgroup $N$ is a normal subgroup of $G$ if $gN{g}^{-1}=N$ for every $g\in G$. Define multiplication on its left cosets byTo check that this is well-defined, replace $g,h$ by $g{n}_1,h{n}_2$ with $n_1,n_2\in N$. Thenbecause normality gives $h^{-1}{n}_{1}h\in N$. Thus the product depends only on the two cosets. Associativity follows from that of $G$, the identity is $N$, and the inverse of $gN$ is $g^{-1}N$. These are all the group axioms, so $G/N$ is a group under coset multiplication. The natural projection $g\mapsto gN$ is a surjective group homomorphism with kernel $N$.

Use additive notation and take $G=\mathbb{Z}/4\mathbb{Z}$, $N=\{0,2\}$. Its quotient group is $C_2$. A group homomorphism from $C_2$ sends its generator to an element annihilated by $2$, so the image must be either $0$ or $2$. Both project to the identity coset $N$, and neither can project to the nonidentity generator of $G/N$. Therefore this quotient projection has no homomorphic section. It is a nonsplit group extension, despite both $G$ and $N$ being nontrivial finite groups.

Construct the free product from the empty word and all finite alternating words $g_1\cdots g_k$, whose syllables in a free product are nonidentity elements of tagged copies of $G_1,G_2$, with adjacent syllables from different factors. Multiply by concatenating, multiplying adjacent elements in the same factor, and deleting identities until the word is reduced. The normal form theorem for a free product gives a unique result; reducing three concatenated words gives the same result under either parenthesization, so multiplication is associative. The empty word is the identity, and the inverse reverses the word and inverts each syllable. Each factor embeds as words of length one.

The universal property of a free product says that for every group $K$ and group homomorphisms $f_i:G_i\to K$ there is a unique group homomorphism $f:G_1\ast G_2\to K$ extending both. Explicitly, send a reduced word to the product of its syllable images; reduction preserves this product. This gives existence, while generation by the two factors gives uniqueness.

A standard form of Klein's combination theorem, or the ping-pong lemma, is the following. Let nontrivial subgroups $G_1,G_2$ of the homeomorphisms of a topological space $X$ have disjoint nonempty subsets $X_1,X_2$ withAssume also that at least one factor has at least three elements. Then the generated subgroup is $\boxed{⟨G_1,G_2⟩\cong G_1\ast G_2}$. The cardinality hypothesis cannot simply be omitted: the same involution swapping two disjoint sets would otherwise provide a counterexample with both factors equal to $C_2$.

Here is the ping-pong lemma proof. A reduced word of odd length begins and ends in the same factor, so repeated application of the displayed inclusions sends the other factor's domain into that factor's domain. Disjointness shows that the word is not the identity. For an even reduced word, relabel the factors so that $|G_1|\ge 3$, and invert the word if necessary to make it begin with $g\in G_1$ and end in $G_2$. Choose $s\in G_1\setminus \{1,g^{-1}\}$. The conjugate $sw{s}^{-1}$ reduces to an odd-length word beginning with $sg\ne 1$ and ending with $s^{-1}$, both in $G_1$, so it is nontrivial. Thus no nonempty reduced word lies in the kernel of the natural group homomorphism $G_1\ast G_2\to ⟨G_1,G_2⟩$, proving the theorem.

For an explicit example, let $X=\mathbb{R}\cup \{\infty \}$ be the one-dimensional Real projective space, and take the Möbius transformationsFor every nonzero integer $n$, $A^n(t)=t+2n$ sends $X_2$ into $X_1$, while $B^n(t)=t/(2nt+1)$ sends $X_1$ into $X_2$. Indeed $|2nt+1|>|t|$ when $|t|>1$, and $B^n(\infty )=1/(2n)$. Both transformations have infinite order. Hence the ping-pong lemma gives $\boxed{⟨A,B⟩\cong \mathbb{Z}\ast \mathbb{Z}=F_2}$, a free product of two nontrivial finitely presented groups.

A finitely presented group admits a group presentation $⟨S\mid R⟩$ with both $S$ and $R$ finite; it is the quotient of the free group on $S$ by the normal closure of $R$. Ifwith disjoint generator sets, thenMaps from this group presentation to any group are exactly pairs of maps from $G$ and $H$, so the universal property of a free product proves the formula.

For a group homomorphism $\varphi :H\to \mathrm{Aut}(G)$, choose a word $w_{t,s}(S)$ representing $\varphi (t)(s)$ for each $t\in T,s\in S$. The presentation of a semidirect product isThe presentation maps onto the specified semidirect product. Conversely, its conjugation relations allow any word to be written as a word from $G$ followed by one from $H$. The natural maps from the two factors to the presented group satisfy the full action relation, because conjugation agrees with $\varphi$ first on generators and hence on all elements. They define the reverse group homomorphism $(g,h)\mapsto gh$. The two maps are inverse on every generator, proving the group isomorphism. There are finitely many cross-relations, so the result is again a finitely presented group.

Apply this to the specified permutation action. The presentation isEliminate $y=cx{c}^{-1}$ and $z=c^{2}x{c}^{-2}$. The last cross-relation becomes $c^{3}x{c}^{-3}=x$, already implied by $c^3=1$. ThusBoth factors are nontrivial finitely presented groups, as required.

Both displayed matrices have determinant one and integer entries, so the generated group $M$ lies in ${\mathrm{SL}}_2(\mathbb{Z})$. For any nonidentity matrix $g\in M$, some entry of $g-I$ is a nonzero integer $k$. Choose a prime number $p$ not dividing $k$. The reduction modulo a prime in an integral matrix group gives a group homomorphismto a finite group in which $g$ is nonidentity. $\boxed{M\text{ is residually finite}.}$ No determination of the abstract subgroup generated by the two matrices is needed.

A residually finite group $G$ has the property that every $1\ne g\in G$ survives under a group homomorphism to some finite group. Equivalently, the intersection of its finite-index normal subgroups is trivial. A Hopfian group is a group for which every surjective endomorphism is an automorphism.

Suppose $G$ is generated by $d$ elements. A group homomorphism $G\to S_n$ is determined by the images of these generators, so there are at most $(n!{)}^d$ such maps. Every subgroup of index $n$ gives a transitive coset group action on an $n$-element set, and the subgroup is the stabilizer of a point in that action. There are at most $n$ point stabilizers per action. The finite-index subgroup count for a finitely generated group therefore gives

Now let $\psi :G\to G$ be a surjective endomorphism. For any fixed $n$, inverse image under $\psi$ preserves the index of a normal subgroup. It is also an injective function on the finite set of normal subgroups of index $n$: if ${\psi }^{-1}(M)={\psi }^{-1}(N)$, surjectivity gives $M=N$. It is therefore a permutation of that finite set. Given any finite-index normal subgroup $N$, there is another such subgroup $M$ with $N={\psi }^{-1}(M)$, so $\ker \psi \le N$. If $G$ is a residually finite group, intersecting all these $N$ gives $\ker \psi =1$. Hence $\psi$ is an automorphism. Every finitely generated group that is a residually finite group is a Hopfian group.

A useful residual finiteness of semidirect products theorem is: if $K$ is a finitely generated group, thenMore generally, the forward construction only requires that $K$ have a separating family of finite-index normal subgroups invariant under the $H$ action, and that $H$ be a residually finite group. Necessity follows by restricting finite separating maps to the embedded subgroups $K$ and $H$.

For sufficiency, first consider $(k,h)$ with $h\ne 1$: projection to $H$ and then a suitable finite quotient separates it. If $h=1$ and $k\ne 1$, choose a finite-index normal subgroup $U⊴K$ with $k\notin U$. Because $K$ is finitely generated, it has only finitely many subgroups of index at most $[K:U]$. Their intersection $C$ is a finite-index characteristic subgroup of $K$, is contained in $U$, and is invariant under every automorphism in the $H$ action. The quotient $K/C$ is finite. Let $\rho :H\to \mathrm{Aut}(K/C)$ be the induced action. The mapis a group homomorphism to a finite group and separates $(k,1)$. This proves the theorem and the more general invariant-subgroup criterion. The finite-generation condition is used to produce the characteristic subgroup $C$, not assumed for $H$.

Let $a,b$ be the generating involutions. If $a\ne b$, set $x=ab$ and $y=b$. Then $x\ne e$, so its order of a group element is at least two or infinite, andTherefore $x,y$ generate the group and satisfy the required dihedral inversion relation. If $a=b$, the group is $C_2$ and $x=y=a$ gives its permitted degenerate description. This proves the two-involution characterization of a dihedral group without excluding either $C_2$ or the infinite dihedral group.