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The map $(a,b)\mapsto \mathrm{sgn}(a)$ is a surjective group homomorphism with kernel $H_0$, so $H_0$ is a normal subgroup and $G/H_0\cong C_2$. The subgroup $H_1$ is not normal: for a translation $t=(1,c)$,which leaves $H_1$ when $a\ne 1$ and $c\ne 0$. Finally, $(a,b)\mapsto a$ is a surjective homomorphism onto ${\mathbb{R}}^{\ast }$ with kernel $H_2$, so $H_2◃G$ and the quotient group $G/H_2\cong {\mathbb{R}}^{\ast }$.

Whenever a combinatorial loop $\alpha$ traverses an oriented edge and immediately traverses the same edge backwards, delete that backtracking pair. Each deletion is a homotopy relative to endpoints inside $Z$ and reduces the edge length by two, so the process terminates at a reduced, hence locally injective, combinatorial loop $\beta$.

The universal cover $\tilde{Y}$ of a connected graph is a tree. The lift $\tilde{\beta }$ is also locally injective because a covering map is a local graph isomorphism. A locally injective edge path in a tree cannot repeat a vertex: the segment between two successive visits would be a nonempty reduced closed path, whereas every closed path in a tree backtracks. Thus $\tilde{\beta }$ is injective unless $\beta$ is constant.

Now let a loop in $Z$ become null-homotopic in $Y$. Its reduced representative $\beta$ lifts to a closed path in $\tilde{Y}$. The preceding injectivity forces that lift, and hence $\beta$, to be constant. The original loop is null-homotopic in $Z$, proving thatis an injective group homomorphism.

Letbe the orientation-preserving $(2,3,7)$ hyperbolic triangle group. DefineEvery defining relator of $G$ maps to the identity, and $A,B,C$ lie in the image, so this is a surjective group homomorphism. Since$\Delta (2,3,7)$ is a non-elementary Fuchsian group.

A group homomorphism from an additive cyclic group ${\mathbb{Z}}_n$ is determined by the image $a$ of $1$, subject exactly to $a^n=e$ in multiplicative notation, or $na=0$ in an additive target.

For ${\mathbb{Z}}_{18}\to {\mathbb{Z}}_{15}$, the relation is $18a\equiv 0(\mathrm{mod}15)$, equivalently $3a\equiv 0(\mathrm{mod}15)$. The possibilities are $a=0,5,10$, and each defines a homomorphism between cyclic groups. Thus there are three such maps, an instance of the number of homomorphisms between finite cyclic groups.

For ${\mathbb{Z}}_9\to S_4$, the image of $1$ must be a permutation whose order of a group element divides $9$. A permutation on four symbols can have order $1,2,3$ or $4$, so the allowed images are the identity and the three-cycles. There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)(3-1)!=8$ three-cycles: choose their three symbols and one of the two cyclic orientations. Hence there are nine maps.

Finally, each transposition in $S_5$ has order $2$, so its image in ${\mathbb{Z}}_5$ must have order dividing both $2$ and $5$, and is therefore the identity. Every permutation is a product of transpositions, so every element must map to the identity. There is one map. In the order asked, the counts are

Let $n=o(g)$. A group homomorphism preserves powers and the identity element, soThe preceding divisibility property givesNow assume $\varphi$ is surjective and choose $h\in H$ of order $m$. Take a preimage $g$ of $h$, and write $o(g)=mr$, as the divisibility result permits. The element $g^r$ has order exactly $m$: its $m$th power is $e$, while $(g^r{)}^s=e$ implies $mr\mid rs$, hence $m\mid s$. Therefore $G$ contains an element of order $m$, namely $g^r$.

The preimage $g$ itself need not have order $m$; taking the suitable power is the essential step. The constructed element need not map to the original $h$. This proves that element orders in a finite quotient occur in the original group.

The group action law givesThus ${\varphi }_g$ is a bijection with inverse ${\varphi }_{g^{-1}}$, hence a permutation of $S$. Also, for every $s$,Therefore the assignment $\Phi (g)={\varphi }_g$ takes values in the symmetric group $\mathrm{Sym}(S)$ and satisfiesIt is a group homomorphism, with the group product on the right interpreted as composition of permutations.

By Lagrange's theorem, the index of a subgroup is $[G:H]=75/25=3$. Thus the set of left cosets has three elements, and the preceding group action yields a group homomorphism $\Phi :G\to S_3$ after labelling them.

Its image is a subgroup of $S_3$, so its order divides $6$. By the first isomorphism theorem, $|\mathrm{im}\Phi |=|G|/|\ker \Phi |$ also divides $75$. Hence its order divides $\gcd (6,75)=3$, giving possibilities $1$ and $3$. The action on cosets is transitive: $g_2{g}_1^{-1}$ sends $g_{1}H$ to $g_{2}H$. It cannot be the trivial action on a three-element set, so the image does not have order $1$. ThereforeIt follows that $|\ker \Phi |=75/3=25$. Part (b)(i) gives $\ker \Phi \le H$, and $|H|=25$; equal finite orders force equality. A kernel of a group homomorphism is a normal subgroup, soThis is an instance of the fact that a subgroup of index the smallest prime divisor is normal.

Define $g\cdot (g_{1}H)=(g{g}_1)H$ on the set of left cosets. It is well-defined: if $g_{1}H=g_{2}H$, then $g_2=g_{1}h$ for some $h\in H$, so $(g{g}_2)H=(g{g}_{1}h)H=(g{g}_1)H$. The identity fixes each coset, and associativity in $G$ givesHence this is a left group action, and part (a) supplies its group homomorphism $\Phi :G\to \mathrm{Sym}(S)$.

An element of the kernel of a group homomorphism $\Phi$ fixes every coset, in particular $H$ itself. Thus $g\in \ker \Phi$ implies $gH=H$, which implies $g\in H$. ThereforeThe subgroup symbol is justified because a kernel of a group homomorphism is a subgroup, and the argument proves containment.

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.