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For $h\in H$, define $\pi (h)(n)=hn{h}^{-1}$. Since $N$ is a normal subgroup, this sends $N$ into itself. It preserves multiplication, and its inverse is conjugation by $h^{-1}$, so it is an group automorphism of $N$. MoreoverThus $\pi :H\to \mathrm{Aut}(N)$ is a group homomorphism, induced by the group action of conjugation.

If $\pi$ is trivial, every element of $H$ commutes with every element of $N$. Henceis a group homomorphism, since $(n_1{h}_1)(n_2{h}_2)=n_1{n}_2{h}_1{h}_2$. Its image of a group homomorphism is a subgroup containing both $N$ and $H$, and every product $nh$ lies in $⟨N,H⟩$, so its image is exactly $⟨N,H⟩$. If $nh=e$, then $n=h^{-1}\in N\cap H=\{e\}$, so $n=h=e$. Thus $F$ is injective and gives

Write $C_p=⟨g⟩$. A group homomorphism from this cyclic group is determined by the image of $g$, because $\varphi (g^j)=\varphi (g{)}^j$. Each choice $\varphi (g)=g^a$, with $a=0,\dots ,p-1$, gives a well-defined group homomorphism. It is an group automorphism exactly when $g^a$ generates $C_p$.

For a prime number $p$, every $a=1,\dots ,p-1$ is coprime to $p$, so the powers of $g^a$ run through all of $C_p$. The choice $a=0$ gives the trivial group homomorphism and fails to be bijective. Consequently

A map $\varphi :G\to G$ is a group isomorphism if it is a bijection and a group homomorphism, meaningSuch a map from a group to itself is an group automorphism. Its inverse is also a group homomorphism: if $u=\varphi (x)$ and $v=\varphi (y)$, then ${\varphi }^{-1}(uv)=xy={\varphi }^{-1}(u){\varphi }^{-1}(v)$.

A left group action is a map $a:G\times X\to X$ satisfyingDefine ${\rho }_a(g):X\to X$ by ${\rho }_a(g)(x)=a(g,x)$. The action axioms giveHence each ${\rho }_a(g)$ is a bijection, with inverse ${\rho }_a(g^{-1})$, and belongs to the symmetric group $\mathrm{Sym}(X)$. Finally,so ${\rho }_a:G\to \mathrm{Sym}(X)$ is a group homomorphism.

For each $g\in G$, left translation in a group $L_g:x\mapsto gx$ is a bijection, with inverse function $L_{g^{-1}}$. Thus $\theta (g)=L_g$ belongs to $\mathrm{Sym}(G)$. For $g,h,x\in G$, associativity of the group operation givesHence $\theta$ is a group homomorphism. If $\theta (g)=\theta (h)$, evaluation at the identity element $e$ gives $g=ge=he=h$. ThereforeThis is the left regular representation used in Cayley's theorem.

Fix $x\in G$. The integer powers of a group element satisfyFor nonnegative integers these rules follow by associativity and counting factors; negative integers are handled by $x^{-r}=(x^{-1}{)}^r$, and opposite signs by cancellation of adjacent inverse elements. All factors are powers of the same element, so no abelian group assumption is involved.

It follows that $z\mapsto \rho (z,x)=x^z$ is a group homomorphism from the additive group of integers to $G$, andThe homomorphism uses ordinary addition on $\mathbb{Z}$; composition of the power functions corresponds to ordinary multiplication of exponents.

Assume $T$ is a group homomorphism. For any $a,b\in G$, expanding $(ab{)}^3=a^3{b}^3$ and cancelling the initial $a$ and final $b$ gives the useful identityWe also need a consequence of compatibility with conjugation. Calculate the cube of $gz{g}^{-1}$ in two ways:Cancelling shows $g^2{z}^3=z^3{g}^2$. Thus the square of every element of $G$ commutes with every element of the image of a group homomorphism $J=T(G)$.

Now take any $x,y\in J$. In particular $x^{2}y=y{x}^2$. Applying (1) to $x,y$ and then using this commutation givesThis proves the abelian image of a cubing homomorphism property. Since the image of a group homomorphism is a subgroup, $\boxed{T(G)\text{ is an abelian group}}$.

Use the symmetric group $S_3$, composing permutations from right to left. Set $x=(12)$ and $y=(23)$. Each is a transposition, so $x^3=x$ and $y^3=y$, whereas $xy=(123)$ has order of a group element three. ConsequentlyThus the cubing map on $S_3$ is not a group homomorphism.

If $T$ is a surjective group homomorphism, then $G=T(G)$, which is an abelian group by the preceding part.

If $T$ is an injective group homomorphism, then for all $a,b\in G$,because $T(a),T(b)$ lie in the abelian group $T(G)$. Injectivity gives $ab=ba$. Hence $\boxed{G\text{ is abelian in either case}}$.

Yes: use the non-abelian group $H$ from part (a). Every element is $I_3+N$ for a strictly upper triangular matrix $N$ with $N^3=0$. In the finite field ${\mathbb{Z}}_3$, the binomial theorem for commuting matrices givesThus $\boxed{T(h)=I_3\text{ for all }h\in H}$. This constant identity-valued map is a group homomorphism, since $T(xy)=I_3=T(x)T(y)$. Part (a)(iv) already proves that $H$ is not an abelian group. Here $T$ is neither injective nor surjective, consistently with the preceding part.

For $n\ge 2$ with $n\ne 6$, bijectivity of $\varphi$ and the previous part implyThe supplied numerical fact forces $k=1$. Thus $\varphi$ permutes the transpositions.

Suppose $n\ge 3$, and write $s_r=\varphi ((1r))$ for $r=2,\dots ,n$. These are distinct transpositions. For $r\ne t$, the product $(1r)(1t)$ has order of a group element three, so $s_r{s}_t$ also has order three. Two distinct transpositions have this property precisely when their two-element sets of moved points meet in one point: disjoint ones have a product of order two, whereas ones sharing one point have a three-cycle as product. The two-element support of a permutation associated with each $s_r$ therefore meets that of every other $s_t$.

We now show that they have a common point. Choose two; the support of a permutation is $\{a,b\}$ for one and $\{a,c\}$ for the other, where $a,b,c$ are distinct. Any further two-element set meeting both either contains $a$ or is $\{b,c\}$. If $\{b,c\}$ occurs, any two-element set meeting all three must itself be one of $\{a,b\},\{a,c\},\{b,c\}$: a set containing $a$ must also contain $b$ or $c$, and a set omitting $a$ must contain both $b,c$. Hence a pairwise-intersecting family with no common point consists of exactly those three edges of a triangle.

For $n\ge 5$, there are $n-1\ge 4$ distinct $s_r$, ruling out that triangle. For $n=3$, the two sets already have a common point. For $n=4$, the triangle alternative would make all three $s_r$ fix the fourth point. They would then generate a proper subgroup of $S_4$, contrary to surjectivity of $\varphi$ and part (a)(iv), which imply that the $s_r$ generate $S_4$. Thus in every case $n\ge 3$ there is a common point $a$, and we may writeThe $b_r$ are distinct and different from $a$, so they exhaust the remaining $n-1$ points. Define a permutation $\sigma$ byThenThe two group homomorphisms $\tau \mapsto \varphi (\tau )$ and $\tau \mapsto {\sigma }^{-1}\tau \sigma$ agree on a generating set of a group, so agree on every product of its elements, hence throughout $S_n$:Finally, $S_1$ is trivial and $S_2$ has a unique nonidentity element, so every automorphism of either is the identity and we can take $\sigma =e$. This completes all degrees $n\ne 6$ and proves that every such automorphism is an inner automorphism.

Representations. A complex group representation is a group homomorphism $\rho :G\to \mathrm{GL}(W)$ to the invertible linear maps on a vector space. It is an irreducible representation when $W\ne 0$ and the only invariant subspaces are $0$ and $W$.

Group table. The supplied matrices satisfy $A=-I$, $C=-B$, and $A^2=B^2=C^2=I$. Their multiplication table isThe table proves closure and gives identity $I$ and an inverse for every element; associativity is inherited from matrix multiplication. The table is symmetric, so the group is abelian. It is the Klein four-group $V\cong C_2\times C_2$.

Dimensions and conjugacy. The sum of squares of irreducible representation dimensions is $|V|=4$. A two-dimensional irreducible would alone contribute four, leaving no room for the ever-present one-dimensional trivial representation. Hence the given two-dimensional representation is reducible. Since the group is abelian, every conjugacy class is a singleton. There are therefore four inequivalent complex irreducible representations, each of dimension one:Taking the trace of the given matrices gives their character of a representation:

Irreducible characters. In a one-dimensional representation, the values on $A$ and $B$ must each be $\pm 1$, since both elements square to $I$. Their values determine the value on $C=AB$. The four choices give the complete character table:Each row is a character because it respects the generating relations, and the rows have unit norm and pairwise zero character inner product. The multiplicities of these rows in $(2,-2,0,0)$ are respectively $0,0,1,1$, by character orthogonality. Thus

Diagonal form. The common invariant lines are spanned by $(1,1{)}^T$ and $(1,-1{)}^T$. Use their normalized vectors as columns ofThen simultaneous diagonalization givesReading the first and second diagonal entries across the four matrices recovers ${\chi }_3$ and ${\chi }_4$, respectively, verifying the decomposition explicitly.

Definitions. A group homomorphism $\phi :G_1\to G_2$ preserves multiplication, $\phi (xy)=\phi (x)\phi (y)$. A group isomorphism is a bijective group homomorphism. The kernel of a group homomorphism isPreserving multiplication implies $\phi (e_1)=e_2$ and $\phi (x^{-1})=\phi (x{)}^{-1}$: the first follows by cancelling $\phi (e_1)$ from its square, and the second follows by applying $\phi$ to $x{x}^{-1}=e_1$.

Kernel subgroup. The identity belongs to $K$. For $x,y\in K$, $\phi (xy)=e_2{e}_2=e_2$, so $xy\in K$. Also $\phi (x^{-1})=e_2^{-1}=e_2$, so $x^{-1}\in K$. Thus $\boxed{K\le G_1}$.

Normality. For $g\in G_1$ and $k\in K$,Hence every $gk{g}^{-1}$ is again in $K$, proving $\boxed{K⊴G_1}$. Applying the same statement with $g^{-1}$ gives equality $gK{g}^{-1}=K$, and therefore $gK=Kg$.

Group of cosets. Define the product on the left cosets byTo check that this does not depend on representatives, replace $a,b$ by $a{k}_1,b{k}_2$. Thenbecause normality puts $b^{-1}{k}_{1}b$ in $K$. Thus the product is well defined. It is associative because multiplication in $G_1$ is associative; its identity is $K=e_{1}K$, and the inverse of $aK$ is $a^{-1}K$. Consequently the cosets form the quotient group $\boxed{G_1/K}$.

The map $aK\mapsto \phi (a)$ is well defined and gives an isomorphism $G_1/K\cong \mathrm{im}\phi$. If the original mapping is onto $G_2$, this becomes $G_1/K\cong G_2$. No surjectivity is needed for the kernel of a group homomorphism to be a normal subgroup or for the quotient group to exist.

Definitions and labels. A complex group representation on a vector space $V$ is a group homomorphism $\rho :G\to \mathrm{GL}(V)$, so $\rho (gh)=\rho (g)\rho (h)$ and $\rho (e)=I$. It is an irreducible representation if $V\ne 0$ and its only invariant subspaces are $0$ and $V$.

Read permutation matrices by their action on column basis vectors. The labels in the original PDF areIn particular, the original PDF gives the last row of $C$ as $(1,0,0)$; the provided TeX transcription's $(0,0,1)$ is incorrect and would not give an invertible permutation matrix.

Multiplication and conjugacy. Matrix products act on columns with the right factor first. For example $A^2=B$, $AD=C$, and $DA=E$. The complete multiplication table, with row factor multiplied by column factor, isConjugating a permutation relabels its cycles. Thus the conjugacy classes areof sizes one, two and three. These are respectively the identity, the three-cycles and the transpositions in the symmetric group on three letters.

Dimensions. The sum of squares of irreducible representation dimensions equals the group order, which here is six. A three-dimensional irreducible representation would already contribute nine, so the displayed representation is reducible. There are three inequivalent complex irreducible representations, one for each conjugacy class, including the trivial representation of dimension one. If their dimensions are $1,d_2,d_3$, then $d_2^2+d_3^2=5$, whose positive integer solution is $\{d_2,d_3\}=\{1,2\}$. HenceOne can also see reducibility directly: every permutation matrix fixes $(1,1,1{)}^T$.

Characters. The character of a representation is the trace of its representing matrix, constant on each conjugacy class. The given permutation representation has character equal to the number of fixed basis vectors:in class order $(I;A,B;C,D,E)$. A trivial representation has character $(1,1,1)$. For the second one-dimensional character, a transposition has value $\pm 1$ because its square is the identity. The two three-cycles are conjugate and inverse, so their common value $a$ obeys both $a^2=1$ and $a^3=1$, forcing $a=1$. Character orthogonality with the trivial character then requires $1+2+3b=0$, giving $b=-1$. This determines the sign representation without constructing its matrices.

Write the two-dimensional character as $(2,x,y)$. Character orthogonality with the first two rows givesso $x=-1$, $y=0$. Its squared norm is $(4+2)/6=1$, as required. The complete character table is thereforeThe last row belongs to the standard representation of the symmetric group.

To decompose the given representation, use the character inner productIts multiplicities areThusConcretely the invariant line is spanned by $(1,1,1{)}^T$ and its invariant complement is $\{(x,y,z):x+y+z=0\}$. Their characters are exactly the first and third rows, agreeing with the decomposition obtained entirely from character orthogonality.

A group homomorphism $h:G\to F$ preserves multiplication: $h(g_1{g}_2)=h(g_1)h(g_2)$. It sends the identity element to the identity and inverse elements to inverse elements. Its group kernel is $K=\{g\in G:h(g)=e_F\}$. This contains $e_G$ and is closed under multiplication and inverse elements, so it is a subgroup. For $k\in K$ and $g\in G$,Thus $gK{g}^{-1}\subseteq K$; applying the same argument with $g^{-1}$ gives equality. Hence the group kernel is a normal subgroup.

Let $S_2=\{e,\tau \}$ with $\tau =(12)$. The parity of a permutation defines the desired map from $S_3$:The sign of a permutation is multiplicative, so parity adds modulo two under composition, exactly matching ${\tau }^2=e$. Thus this is a group homomorphism; it is surjective because a transposition maps to $\tau$. Its group kernel isThis is the alternating group on three objects, a normal subgroup of index two in $S_3$.

Let $r$ be a quarter-turn and $s$ a reflection of the square. The dihedral group is generated by $r,s$ with $r^4=s^2=1$ and $srs=r^{-1}$, and its elements are $1,r,r^2,r^3,s,sr,s{r}^2,s{r}^3$. Two generators are minimal because this group is not cyclic. Its subgroups of order two are $⟨r^2⟩$ and $⟨s{r}^j⟩$ for $j=0,1,2,3$. Its subgroups of order four are $⟨r⟩$, $⟨r^2,s⟩$ and $⟨r^2,sr⟩$.

A group homomorphism obeys $\varphi (gh)=\varphi (g)\varphi (h)$ for all elements; a group isomorphism is a bijective group homomorphism. Take the rotation subgroup $⟨r⟩$ and defineThe quaternion element $i$ has order four, so the map is well defined modulo four, preserves products and is injective onto $\{1,i,-1,-i\}$. Surjectivity onto all eight quaternion elements is not required of a homomorphism.

A cyclic group is generated by the powers of one element. The final claim needs the usual qualification no nontrivial proper subgroup, and $|G|>1$: the trivial subgroup is proper in every nontrivial group, even one of prime order. With that interpretation, if $G=⟨g⟩$ has prime order $p$, Lagrange's theorem permits only subgroup orders one and $p$. Conversely, if its order $n>1$ is composite, choose a divisor $d$ with $1<d<n$. The subgroup $⟨g^d⟩$ has order $n/d$, strictly between one and $n$. This proves the assertion described by groups with no proper nontrivial subgroups.

A group homomorphism preserves multiplication, $\varphi (ab)=\varphi (a)\varphi (b)$. A group isomorphism is a bijective homomorphism. An isomorphism from a group to itself is a group automorphism.

The automorphism group contains the identity map. Composition is associative, and if $\varphi ,\psi$ are automorphisms, $(\varphi \circ \psi )(ab)=\varphi (\psi (a)\psi (b))=(\varphi \circ \psi )(a)(\varphi \circ \psi )(b)$; composition is also bijective. For $u=\varphi (a)$ and $v=\varphi (b)$, ${\varphi }^{-1}(uv)={\varphi }^{-1}(\varphi (ab))=ab={\varphi }^{-1}(u){\varphi }^{-1}(v)$, proving the inverse bijection is a homomorphism. These facts establish all group axioms.

A finite cyclic group is $G=⟨g⟩=\{e,g,\dots ,g^{n-1}\}$ with $g$ of order $n$. The map $g^k\mapsto e^{2\pi ik/n}$ is well defined modulo $n$, bijective onto the roots of unity, and preserves multiplication. It identifies $G$ with the multiplicative group of $n$th roots of unity.

For prime $n$, Lagrange's theorem leaves only subgroup orders one and $n$: there is no nontrivial proper subgroup. The trivial subgroup $\{e\}$ is nevertheless proper when $n>1$, so that qualification is necessary in the printed claim. The characterization of groups with no proper nontrivial subgroups makes the distinction explicit.

For composite $n$, choose a prime divisor $p$ and take $\boxed{H_0=\{e\},H_1=⟨g^p⟩}$, of orders $1$ and $n/p$. They are distinct proper subgroups, with $H_1$ nontrivial. If the question intended two nontrivial proper subgroups, that demand is not always possible: a cyclic group of order $p^2$ has only the one of order $p$. The subgroups and quotients of a cyclic group are controlled by its divisors. The order-one group is a separate trivial case, with no proper subgroups.

A normal subgroup $H$ satisfies $gH{g}^{-1}=H$ for every $g\in G$. A group homomorphism preserves multiplication, $\Phi (ab)=\Phi (a)\Phi (b)$. It follows that $\Phi (e_1)=e_2$ and $\Phi (g^{-1})=\Phi (g{)}^{-1}$.

The kernel of a group homomorphism $K=\{g:\Phi (g)=e_2\}$ contains the identity and satisfies $a{b}^{-1}\in K$ for $a,b\in K$, so is a subgroup. For $k\in K$, $\Phi (gk{g}^{-1})=\Phi (g)e_2\Phi (g{)}^{-1}=e_2$. Thus $gK{g}^{-1}\subseteq K$, and applying the same inclusion to $g^{-1}$ gives equality. Hence the kernel is a normal subgroup.

The product $f={\prod }_{i<j}(x_i-x_j)$ is a Vandermonde determinant up to a fixed overall sign. A transposition exchanging $x_p,x_q$ reverses their mutual factor. Every other factor involving one of these variables is paired with the corresponding factor involving the other; those paired products are unchanged. ThereforeIf a permutation has decompositions into $r$ and $s$ transpositions, they act as $(-1{)}^{r}f$ and $(-1{)}^{s}f$. This polynomial is nonzero, for example at pairwise distinct real inputs, so the signs must agree. Thus all decompositions have the same parity, and an even decomposition forces every decomposition to be even. This defines the sign of a permutation; its kernel is the alternating group, linking the parity argument to the normal-kernel result above.

A real two-dimensional group representation is a group homomorphism $D:G\to GL(2,\mathbb{R})$, so $D(AB)=D(A)D(B)$ and $D(I)=I$.

Substitute $\tilde{x}=\alpha x+\beta y$ and $\tilde{y}=\gamma x+\delta y$ into the quadratic expression. Its coefficients of $x^2$ and $y^2$ are ${\alpha }^2\hat{a}+{\gamma }^2\hat{b}$ and ${\beta }^2\hat{a}+{\delta }^2\hat{b}$. The cross term is zero for every matrix in this particular cyclic group, because each has one nonzero entry per row. ThusLet $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ generate the group. For $A=\pm I$, $M=I$; for $A=\pm J$, $M=S:=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Since $S^2=I$, each $M$ is its own inverse and $D(J^k)=S^k$. The group law is respected: $S^{k+l}=S^k{S}^l$, and $S^4=I$ makes it independent of the representative modulo four. This proves that $D=M^{-1}$ is a group representation.

This quarter-turn action on diagonal quadratic forms hasso it is not faithful. Quadratic forms cannot distinguish a simultaneous sign reversal of both coordinates. The invariant coefficient directions $(1,1)$ and $(1,-1)$ correspond to the fixed form $x^2+y^2$ and the form $x^2-y^2$ whose sign changes under a quarter-turn.

A group homomorphism satisfies $\phi (g_1{g}_2)=\phi (g_1)\phi (g_2)$; a bijective one is a group isomorphism. Its kernel of a group homomorphism is $K=\{g:\phi (g)=1_H\}$. It is a subgroup, and for $k\in K$, $g\in G$,Thus $gK{g}^{-1}\subseteq K$, and applying the same argument to $g^{-1}$ gives equality. Therefore $K$ is a normal subgroup.

Write $B_n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$. Direct matrix multiplication gives $B_n{B}_m=B_{n+m}$, $B_0=I$ and $B_n^{-1}=B_{-n}$. Consequently these matrices form the integer shear group, a subgroup of $GL(2,\mathbb{R})$. It is an infinite cyclic group, generated by $B_1$, since $B_1^n=B_n$ and no nonzero integer power is the identity.

The map $n\mapsto B_n$ is a bijective group homomorphism from the additive group of integers, proving the required group isomorphism. There are proper subgroups: for example $\{B_{2n}:n\in \mathbb{Z}\}$ omits $B_1$. In fact every subgroup is $\{B_{mn}:n\in \mathbb{Z}\}$ for an integer $m\ge 0$, by the classification of subgroups of an infinite cyclic group; $m=0$ gives the trivial subgroup and $m=1$ the whole group.