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The identity belongs to the center of a group. If $z,w$ commute with every $g$, then $zw$ does too; rearranging $gz=zg$ also shows that $z^{-1}$ commutes with $g$. Hence $Z(G)$ is a subgroup. Moreover $gz{g}^{-1}=z$ for $z\in Z(G)$, so it is a normal subgroup.

Suppose $|G|=p^n>1$. The conjugation action partitions $G$ into conjugacy classes. The orbit-stabilizer theorem gives class size $[G:C_G(x)]$, where $C_G(x)$ is the centralizer of $x$. By Lagrange's theorem, this size is a power of $p$; it is one exactly when $x$ is central. Thus the class equation has the formIt follows that $p\mid |Z(G)|$. Since the identity is central, $|Z(G)|\ge 1$, and thereforeThis is the nontrivial-centre property of a finite p-group.

Lagrange's theorem states that, for a finite group $G$ and a subgroup $H$,To prove it, define an equivalence relation by $x\sim y$ when $x^{-1}y\in H$. Its classes are the left cosets $xH$, which therefore partition $G$. Left multiplication by $x$ gives a bijection $H\to xH$, so every coset has $|H|$ elements. Counting this partition gives the formula; in particular $|H|$ divides $|G|$. Applying the theorem to the cyclic subgroup generated by an element shows that its order of a group element divides $|G|$.

Write the given order-two normal subgroup as $N=\{1,z\}$. Conjugation must preserve $N$ and cannot send its nonidentity element to the identity, so $gz{g}^{-1}=z$ for all $g\in G$. Thus $z$ is in the center of a group: a normal subgroup of order two is central.

Choose $g\notin N$. By Lagrange's theorem, its order of a group element is $2$, $p$ or $2p$. It cannot have order two: since $z$ is central, the four distinct elements $1,z,g,zg$ would form a subgroup of order four, whereas $4$ does not divide $2p$ for odd $p$. If $g$ has order $2p$, it already generates $G$. Otherwise $g$ has order $p$, commutes with $z$, and $(gz{)}^k=1$ implies $g^k=z^{-k}\in ⟨g⟩\cap N=\{1\}$. Hence both $p$ and $2$ divide $k$, so $gz$ has order $2p$. In either case

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

In an irreducible complex group representation $\rho$, every $\rho (z)$ for $z$ in the center of a group commutes with the entire representation. Schur lemma therefore gives $\rho (z)={\lambda }_{z}I$. Faithfulness makes $z\mapsto {\lambda }_z$ an injective homomorphism $Z(G)\to {\mathbb{C}}^{\times }$.

Its image is a finite subgroup. If its order is $m$, Lagrange's theorem for finite groups shows that every element is an $m$th root of unity. These roots form a cyclic group, and every subgroup of a cyclic group is cyclic. HenceBoth irreducibility over the algebraically closed field and faithfulness are used: scalar action alone would not embed the centre.

Use the presentation $D_6=⟨r,s:r^3=s^2=1,srs=r^{-1}⟩$. Its subgroup $⟨r⟩$ is the unique subgroup of order $3$, hence is a characteristic subgroup. An automorphism must send $r$ to $r$ or $r^{-1}$ and $s$ to one of the three reflections $r^{j}s$. Thus there are at most six automorphisms.

The center of a group of $D_6$ is trivial, so conjugation embeds $D_6$ as a group of six inner automorphisms. These exhaust the possibilities. Hence every automorphism of $D_6$ is inner.

Fix $\epsilon >0$ and set $d=2/\epsilon$. Suppose, towards a contradiction, thatwhere $\lambda$ will absorb constants depending only on $\epsilon$. Put $n=\lfloor |G{|}^{\epsilon }\rfloor$. Once $\lambda$ is large enough, $n\ge N(d)$, $n<D$, andPart (a) yields $H⊴C<G$ with $[G:C]=O_{\epsilon }(1)$, $H\subseteq S^{\lfloor n/2\rfloor }$, and $C/H$ nilpotent of class $O_{\epsilon }(1)$.

The subgroup core ${\bigcap }_{g\in G}gC{g}^{-1}$ is normal in $G$ and has index at most $[G:C]!$. Since $G$ is a simple group, the core is either $\{1\}$ or $G$. In the first case $|G|\le [G:C]!=O_{\epsilon }(1)$, which is excluded by increasing $\lambda$. Hence the core is $G$, so $C=G$.

Now $H⊴G$. Since $n/2<D$, the ball $S^{\lfloor n/2\rfloor }$ is not all of $G$, so $H\ne G$. Simplicity gives $H=\{1\}$, and therefore $G=C/H$ is a nilpotent group. A nontrivial finite nilpotent group has nontrivial center of a group; simplicity would force that center to be all of $G$, making $G$ Abelian. This contradicts the assumption that $G$ is non-abelian. Consequently

Take $X=S^1\vee S^1$, the wedge of two circles, and label its oriented loops $a$ and $b$. For a fixed $n\ge 3$, letbe permutations of the fibre $\{1,\dots ,n\}$. Construct a labelled graph ${\hat{X}}_n$ with these $n$ vertices, an oriented $a$-edge from $i$ to $\alpha (i)$, and an oriented $b$-edge from $i$ to $\beta (i)$. Map every vertex to the wedge point and each labelled edge homeomorphically onto its corresponding loop. Every vertex has exactly one incoming and one outgoing edge of each label, so this is an $n$-sheeted permutation covering of a wedge of circlesIt is connected because the $n$-cycle $\alpha$ acts transitively on the fibre.

A deck transformation induces a permutation $\gamma$ of the vertices that commutes with both monodromy permutations $\alpha$ and $\beta$. Conversely, such a permutation determines a deck transformation, by the deck transformation group as a monodromy centralizer. The permutations $\alpha$ and $\beta$ generate $S_n$: the conjugates ${\alpha }^j\beta {\alpha }^{-j}$ include the adjacent transpositions, which generate the symmetric group. Hence $\gamma$ centralizes $S_n$ and lies in its center of a group. For $n\ge 3$, the center of $S_n$ is trivial, so $\gamma =1$. A deck transformation fixing one point is the identity by part (a), and therefore

No. The center of a group satisfiesbecause $Z(D_8)=\{1,r^2\}\cong C_2$. This centre is not cyclic. The cyclic-center obstruction to a faithful irreducible representation therefore rules out a faithful irreducible complex representation of $C_2\times D_8$.

A group action of $G$ on $X$ is a map $(g,x)\mapsto gx$ such that $ex=x$ and $(gh)x=g(hx)$. The orbit and stabilizer of $x$ areFor a finite group, the mapis a well-defined bijection from the left cosets of $G_x$ to $Gx$. Each coset has $|G_x|$ elements, so the orbit-stabilizer theorem is

The Cauchy theorem for groups says that if a prime number $p$ divides $|G|$, then $G$ has an element of order $p$. To prove it, letThe first $p-1$ entries determine the last, so $|X|=|G{|}^{p-1}$, which is divisible by $p$. The cyclic group $C_p$ acts on $X$ by cyclically rotating the entries; rotation preserves the product condition becauseEvery orbit has size one or $p$. The fixed points are exactly the tuples $(g,\dots ,g)$ with $g^p=e$. Their number is therefore divisible by $p$. Since the identity gives one fixed point, there is another, and its entry has order $p$.

Now let $|G|=33$. Cauchy's theorem gives a subgroup $H$ of order $11$. Let $H$ act by the conjugation action on the set $\mathcal{X}$ of subgroups of order $11$. It fixes $H$. If it also fixed $K\ne H$, then $H$ would normalize $K$; since $H\cap K=\{e\}$, the product $HK$ would be a subgroup of order $121$, which is impossible. Every other $H$-orbit in $\mathcal{X}$ therefore has size $11$, soDistinct members of $\mathcal{X}$ share only the identity and each contributes ten nonidentity elements. Hence $1+10|\mathcal{X}|\le 33$, forcing $|\mathcal{X}|=1$. Thus $H$ is a normal subgroup.

Conjugation now defines a group homomorphismBecause $H\cong C_{11}$, its automorphism group has order $10$. By the Lagrange theorem, the image has order dividing both $33$ and $10$, so the image is trivial and $H$ lies in the center of $G$. Cauchy's theorem also supplies $x$ of order $3$. If $h$ generates $H$, then $h$ and $x$ commute and $hx$ has order $\mathrm{lcm}(11,3)=33$. Therefore

Schur lemma says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A continuous representation of a topological group is a continuous homomorphismfor a finite-dimensional complex vector space $V$. It is a unitary representation if $V$ has a positive-definite Hermitian inner product for whichfor every $g\in G$ and $v,w\in V$.

For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized Haar measure:Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving unitarity. Since $S^1$ is abelian, the operators $\rho (z)$ commute; since they are unitary, they are normal. Simultaneous diagonalization therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every representation of the circle group is a direct sum of one-dimensional representations.

WriteThe group law and inverse areA calculation givesEvery element of the centre $Z$ occurs by taking, for example, $x^{\prime }=y=0$ and $y^{\prime }=1$. Hence the commutator subgroup is exactly $Z$. The image of a one-dimensional representation is abelian, so the one-dimensional representation kills the commutator subgroup and its kernel contains $Z$.

Now let $(\rho ,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroupis a representation of the circle group. Decompose it into its distinct weight spaces:Because $Z/Z_0$ is central, every $\rho (g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the central circle weight-space decomposition.

Let $n_i=\dim V_i$. The mapis a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value isso ${\theta }_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every ${\theta }_i$ is trivial. Since the ${\theta }_i$ were distinct, $d=1$ and ${\theta }_1=1$.

It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is faithful. This is the real Heisenberg quotient has no faithful finite-dimensional representation.

The equality $x=a^2$ and the relator $[b,x]=1$ show that $x$ commutes with both $a$ and $b$. Since $c=(ab{)}^{-1}$, it also commutes with $c$, and then with $y=b^3$. Hence $x\in Z(G)$. Similarly, $y=b^3$ commutes with $b$ and, by $[a,y]=1$, with $a$; it therefore commutes with $c$ and $x$. Thus

Quotienting by $⟨x,y⟩$ givesA non-elementary Fuchsian group has trivial center: two hyperbolic elements with different pairs of boundary fixed points have only the identity in their common centralizer in ${\mathrm{PSL}}_2(\mathbb{R})$. Therefore the image in the quotient of every element of $Z(G)$ is trivial. It follows that

If $g\in Z(G)$, it commutes with each $x\in G$, so $x^{-1}gx=g$. Its conjugacy class is therefore $C(g)=\{g\}$.

Conversely, if $C(g)=\{g\}$, then $x^{-1}gx=g$ for every $x$. Multiplication by $x$ on the left gives $gx=xg$ for every $x$, which is exactly membership in the center of a group. Thus

The center of a group consists of its elements commuting with every element. First check it is a subgroup. The identity element is central. If $a,b\in Z(G)$ and $x\in G$, then $b^{-1}$ commutes with $x$ andThus $a{b}^{-1}\in Z(G)$, proving closure under the subgroup criterion. Moreover, if $z\in Z(G)$, then $xz{x}^{-1}=z$ for every $x\in G$. Hence conjugation fixes the center elementwise, soTherefore the center of a group is a normal subgroup.

Suppose $\sigma \ne e$. Choose a symbol $i$ moved by $\sigma$ and write $j=\sigma (i)\ne i$. Since $n\ge 3$, choose a third symbol $k$ distinct from $i,j$. Let $\tau =(jk)$, a transposition that fixes $i$. ThenThus ${\tau }^{-1}\sigma \tau \ne \sigma$, proving that the conjugacy class of any nonidentity permutation has more than one element. By part (a)(iii), no such permutation lies in the center of a group. The identity always does, soThe restriction on $n$ is necessary: $S_2$ is an abelian group and equals its own center.

Suppose the quotient group $G/A$ is a cyclic group, generated by $gA$. For any $x,y\in G$, there are integers $r,s$ and $a,b\in A$ such thatSince $a,b$ lie in the center of a group, they commute with every element, in particular with each other and with the integer powers of a group element $g$. ThereforeAs $x,y$ were arbitrary, $G$ is an abelian group. This is the cyclic quotient by the center.

The specified set $A$ is the center of a group $Z(G)$. It contains $e$. If $a,b\in A$ and $y\in G$, thenso $ab\in A$. From $ay=ya$, multiplying on the left and right by $a^{-1}$ gives $y{a}^{-1}=a^{-1}y$, so $a^{-1}\in A$. Hence $A$ is a subgroup.

For $g\in G$ and $a\in A$, the defining commutativity gives $ga{g}^{-1}=ag{g}^{-1}=a$. Thus the image of each element of $A$ under conjugation is in $A$, and indeed $gA{g}^{-1}=A$. Consequently $A=Z(G)$ is a normal subgroup of $G$.

If $G$ is not Abelian, the contrapositive of ii says that $G/A$ is not a cyclic group. The quotient group is finite because $G$ is finite. By iii, every group with fewer than four elements is a cyclic group. HenceEquivalently, the center of a group satisfies $|A|\le |G|/4$, by the coset count in iii.

A conjugacy class $C(x)$ is a singleton exactly when $gx{g}^{-1}=x$ for every $g\in G$, equivalently when $gx=xg$ for every $g$. Thus the singleton conjugacy classes correspond exactly to elements of the center of a group $A$, and there are $|A|$ of them.

Every other conjugacy class has at least two elements. Since the conjugacy classes form a partition of a set $G$, if $r$ is the number of these remaining classes then $2r\le |G|-|A|$. ConsequentlyThis conjugacy class count bound from the center needs no classification of the finite group.

Two elements are conjugate group elements if $h=xg{x}^{-1}$ for some $x\in G$. The conjugacy class of $g$ is $\{xg{x}^{-1}:x\in G\}$. Conjugacy is an equivalence relation: it is reflexive using $x=e$, symmetric by replacing $x$ with $x^{-1}$, and transitive becauseThus the classes partition the group: every element belongs to its own class, and any two classes that intersect are equal. This proves that every element belongs to exactly one class.

If $G$ is Abelian, then $xg{x}^{-1}=g$, so every class is a singleton. In general the elements whose classes are singletons are precisely those commuting with every group element. Thus the center of a group isIt contains $e$. If $a,b\in Z(G)$, then $a{b}^{-1}$ commutes with every $g$, since $b^{-1}$ does and $a$ does. The subgroup test therefore proves that $Z(G)$ is a subgroup. Any two of its elements commute with each other, so it is an abelian subgroup.

For the square centred at the origin with vertices $(\pm 1,\pm 1)$, putThe dihedral group $D_4$ consists of $R^j$ and $R^{j}S$, $j=0,1,2,3$. Its eight actions and their matrices areThese exhaust the symmetries: the image of one vertex has four choices and the image of an adjacent vertex has two choices, determining the entire square. For an explicit pair of nonconjugate elements take $R^2$ and $S$. Both have order two, but their determinants are $+1$ and $-1$. A similarity transformation preserves the determinant, so $R^2$ and $S$ are not conjugate.

To find the centre, use $SR=R^{-1}S$. A rotation $R^j$ commutes with $S$ exactly when $R^j=R^{-j}$, so $j$ is even. No reflection $R^{j}S$ commutes with $R$, since $(R^{j}S)R=R^{j-1}S$ while $R(R^{j}S)=R^{j+1}S$, and $R^2\ne I$. Therefore

The group order $|G|$ is the number of elements of the finite group. A normal subgroup $H\le G$ satisfies $gH{g}^{-1}=H$ for every $g\in G$, equivalently its left and right cosets agree.

For the square, let $R$ be counterclockwise rotation by $\pi /2$, and $s$ reflection in the horizontal axis. The dihedral group isThe first four elements rotate by $0,90,180,270$ degrees. The other four are Euclidean reflections in the horizontal axis, the line $y=x$, the vertical axis and the line $y=-x$, respectively, for a square centred at the origin with sides parallel to the axes.

The eight nontrivial proper subgroups areTo check completeness, Lagrange's theorem allows proper orders one, two and four. The order-two subgroups are generated by the five involutions displayed. An order-four subgroup containing $R$ or $R^3$ is the rotation subgroup. Otherwise it must contain reflections; their product is a rotation, which must be $R^2$, so its reflections differ by that half-turn, giving the two listed Klein subgroups. The trivial subgroup $\{I\}$ is an additional proper subgroup. Thus the PDF's count of eight uses “proper” to exclude the trivial subgroup; with the usual inclusive definition there are nine proper subgroups.

A normal subgroup of order two is central: conjugation fixes its identity and must also fix its only other member. The half-turn $R^2$ commutes with every square symmetry, whereas the reflection subgroups are exchanged by conjugation. Hence the unique order-two normal subgroup isFor the general polygon, write $D_n=\{R^j,R^{j}s:0\le j<n\}$ with $R^n=s^2=I$ and $sRs=R^{-1}$. A rotation $R^j$ commutes with $s$ exactly when $R^j=R^{-j}$, or $2j\equiv 0(\mathrm{mod}n)$. A reflection cannot commute with $R$ when $n\ge 3$, becauseTherefore the center of a group here is $\{I\}$ for odd $n$, and $\{I,R^{n/2}\}$ for even $n$. The order-two normal subgroups correspond exactly to nonidentity central involutions, provingThis agrees with the involutions in an even dihedral group and uses the convention $|D_n|=2n$ throughout.