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A cyclic group of order $p$ contains an element of order two if and only if $2$ divides $p$. Necessity follows from Lagrange's theorem; for even $p$, the unique such element is $c=g_1^{p/2}$. HenceBecause $g_1$ has order $p>2$ and the representation is faithful, $D(g_1)$ cannot be a reflection, which has order two. It is therefore a rotation through a primitive angle $\vartheta =2\pi q/p$, with $\gcd (q,p)=1$. The condition on the greatest common divisor is exactly what makes its order $p$ rather than a proper divisor. Since $p$ is even, $q$ is odd, so $D(c)$ is the half-turn:The character is the trace:Choosing the primitive generator with $q=1$ gives the customary rotation representation. The general coprime value is included because faithfulness alone does not select that generator. These are the two-dimensional cases in the irreducible real representations of a finite cyclic group.

The quotient group $G/N$ consists of the cosets $gN$, with multiplicationFor well-definedness, replacing representatives by $g{n}_1$ and $h{n}_2$ changes their product to $gh(h^{-1}{n}_{1}h)n_2$. Since $N$ is a normal subgroup, the final two factors lie in $N$, so the product coset does not change.

Associativity follows from that in $G$; the identity element is the coset $N$, and $(gN{)}^{-1}=g^{-1}N$. Hence these cosets form a group. By Lagrange's theorem, every subgroup $H\le G$ has order dividing $|G|$, and its subgroup index is $[G:H]=|G|/|H|$. In particularNormality is needed for the quotient multiplication, whereas Lagrange’s counting formula holds for every subgroup.

Let $H$ be a proper subgroup. By Lagrange's theorem, $|H|$ divides $21=3\cdot 7$. Since $H\ne G$, its possibilities areThe subgroup of order one is generated by the identity and is cyclic. For either prime order $p=3$ or $7$, choose $h\in H$ other than the identity. Its order of a group element divides $p$ and is not one, so it is $p$. Thus its powers exhaust $H$, proving $H=⟨h⟩$.

ThereforeThis uses the fact that a group of prime order is cyclic. It does not require the order-21 group itself to be Abelian or cyclic.

A member of $H$ is $A(t,b,t)$. The multiplication law from part (a) becomesIt stays in $H$ and is unchanged when the two factors are exchanged, since $ts=st$. The identity lies in $H$, and $A(t,b,t{)}^{-1}=A(-t,t^2-b,-t)$ also lies in $H$. Thus $H$ is an Abelian subgroup, namely the Abelian plane subgroup of a finite Heisenberg group in this coordinate convention.

There are five choices each for $t,b$, soThis verifies Lagrange's theorem: the subgroup order divides the group order, and the number of distinct left cosets is five.

Explicitly, the map $\phi (A(a,b,c))=a-c$ is a surjective group homomorphism to the additive group of ${\mathbb{F}}_5$, by the product formula. Its kernel of a group homomorphism is $H$. Thus $H$ is even a normal subgroup; the five left cosets are the level sets $a-c=d$ for $d=0,1,2,3,4$, with representatives $A(d,0,0)$. Each contains 25 elements, providing a direct count as well.

Squaring with the product law givesEquality to the identity matrix requires $2a=2c=0$ in ${\mathbb{F}}_5$. Since 2 has a multiplicative inverse modulo 5, this forces $a=c=0$. The middle condition then gives $2b=0$, hence $b=0$. ThereforeThis agrees with Lagrange's theorem: a group of odd order has no element of order two.

The subset specified by $b\ne 0$ is not a subgroup. It omits the identity element. It also fails closure: both $A(0,1,0)$ and $A(0,4,0)$ lie in it, but their product is $A(0,0,0)=I$, whose $b$ coordinate is zero.

A group is a set $G$ equipped with a binary operation satisfying the group axioms: closure; associativity, $(gh)k=g(hk)$; an identity element $e$ with $eg=ge=g$; and an inverse element $g^{-1}$ for every $g$, with $g{g}^{-1}=g^{-1}g=e$.

A subgroup $H$ is a subset that is a group under the restricted operation. Equivalently, it is nonempty and contains $x{y}^{-1}$ whenever $x,y\in H$. A generating set of a group is a set whose elements and inverses can be multiplied in finite words to produce every group element.

The group order is its number of elements, $|G|$. The order of a group element $g$ is the least positive integer $r$ with $g^r=e$, or infinity if there is none. Lagrange's theorem states that for a finite group and subgroup $H$, $|H|$ divides $|G|$ and $|G|=[G:H]|H|$. In particular, every finite element order divides the group order.

A cyclic group is generated by one element, so every element is a power of that generator. A group isomorphism is a bijective group homomorphism: a bijection $\varphi :G\to G^{\prime }$ satisfying $\varphi (gh)=\varphi (g)\varphi (h)$ for all $g,h$.

By Lagrange's theorem, the possible values of the order of a group element areThese are possibilities across groups of order eight; they need not all occur in one group.

If every element has order at most two, every element equals its own inverse. Thus for any $x,y$,so $G$ is an abelian group. Choose $a\ne e$, and then $b\notin ⟨a⟩$. The four elements $e,a,b,ab$ are distinct and form a subgroup $H$: commutativity and $a^2=b^2=e$ prove closure and inverses. Choose $c\notin H$. The elements of $H$ and $Hc$ are disjoint, since $hc\in H$ would imply $c\in H$. These two four-element sets therefore exhaust $G$.

Every element consequently has a unique expression $a^i{b}^j{c}^k$, with $i,j,k\in \{0,1\}$. The map from triples to these products is bijective and respects multiplication by addition of exponents modulo two. Hencea direct product of groups and an elementary abelian group.

For each $g$, the mapis surjective by definition of a coset and injective by left cancellation. Its inverse is multiplication by $g^{-1}$. ThereforeThe same argument with right multiplication proves the analogous result for right cosets. This equality of sizes is the counting step in Lagrange's theorem.

Write $G=\{e,a,b,c\}$. By Lagrange's theorem, the order of a group element divides four. Since no element has order four, all three nonidentity elements have order two. Thus every element is its own inverse. For any $x,y\in G$,so the group is Abelian. Also $ab$ cannot equal $e,a$ or $b$, by cancellation and the fact that $a,b$ are distinct nonidentity elements. Hence $ab=c$. The remaining products are forced, givingThe map $e\mapsto (0,0)$, $a\mapsto (1,0)$, $b\mapsto (0,1)$, $c\mapsto (1,1)$ preserves this table under componentwise addition modulo two. Thereforethe Klein four-group, also called the Vierergruppe.

A cyclic group is generated by one element: $G=⟨g⟩=\{g^n:n\in \mathbb{Z}\}$. The hypothesis needs the qualification no proper nontrivial subgroup: every nontrivial group has the proper trivial subgroup $\{e\}$, so the literal wording admits no examples.

Under that intended hypothesis, choose $g\ne e$. The subgroup $⟨g⟩$ is nontrivial and must therefore be all of $G$, proving cyclicity. If $g$ had infinite order, then $⟨g^2⟩$ would be nontrivial and proper, since $g$ is not an even power of an infinite-order generator. Thus $g$ has finite order $N>1$ and $|G|=N$.

If $N$ were composite, choose a divisor $d$ with $1<d<N$. Then $⟨g^d⟩$ would have order $N/d$, strictly between one and $N$, again contradicting the hypothesis. ConsequentlyConversely, Lagrange's theorem shows that a group of prime order has only itself and its trivial subgroup. This proves the characterization of groups with no proper nontrivial subgroups, even without assuming finiteness at the outset.

The complete subgroup list isThus there are ten subgroups: one trivial, five of order two, three of order four, and the whole group. The five order-two subgroups are mutually isomorphic. The two noncyclic order-four subgroups are mutually isomorphic Klein four-groups; neither is isomorphic to the cyclic group of order four.

For completeness, Lagrange's theorem permits only orders $1,2,4,8$. An order-four subgroup containing $U$ or $U^3$ must be $⟨U⟩$. Otherwise it can contain only $I,U^2$ and elements $U^{j}V$. The product of two distinct such elements is $U^{j-k}$. To avoid generating $U$ or $U^3$, their exponents must differ by two, giving exactly the two listed Klein four-groups. The order-two subgroups come from the five involutions $U^2$ and $U^{j}V$. This proves completeness.

For a normal subgroup $H$, the quotient group consists of its cosets with multiplicationTo check that this is well-defined, replace $g,k$ by $g{h}_1,k{h}_2$ with $h_1,h_2\in H$. Their product issince normality puts $k^{-1}{h}_{1}k$ in $H$. Hence the product does not depend on representatives. Associativity follows from that in $G$, the identity element is $H=eH$, and $(gH{)}^{-1}=g^{-1}H$. These verify the group axioms.

Lagrange's theorem counts the equal-sized cosets:Finally, if $|G|=p$ is prime, choose $g\ne e$. Its cyclic subgroup has order greater than one dividing $p$, hence equals $G$. All elements are powers of $g$, so $g^j{g}^k=g^{j+k}=g^k{g}^j$. Thus

Distinguish the quaternion group generators $\mathbf{i},\mathbf{j},\mathbf{k}$ from the complex scalar $i$. Denote their proposed image matrices byDirect multiplication givesFor example $AB=\mathrm{diag}(i,-i)=C$ and $ABC=C^2=-I_2$. Sending $I$ to $I_2$ and each negative quaternion element to the negative of its positive image now reproduces every quaternion product. The matrices all have determinant one and are invertible, so this is a group representation in $GL(2,\mathbb{C})$.

The eight matrices $\pm I_2,\pm A,\pm B,\pm C$ are distinct. Therefore the group homomorphism is injective, and the representation is faithful; equivalently its kernel is $\{I\}$.

The character of a representation is the trace of its representing matrix. ConsequentlyThese are constant on the five conjugacy classes $\{I\}$, $\{-I\}$, $\{\pm \mathbf{i}\}$, $\{\pm \mathbf{j}\}$, $\{\pm \mathbf{k}\}$.

The complete list of subgroups isTo see completeness, Lagrange's theorem allows only orders $1,2,4,8$. The only element of order two is $-I$, which fixes the order-two subgroup. Every noncentral element has order four and generates the corresponding listed cyclic subgroup. Any order-four subgroup contains such an element, because there are only two central elements; it must equal the cyclic subgroup generated by that element. This accounts for all possibilities.

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$,The left cosets of $H$ partition $G$, and multiplication by a coset representative is a bijection from $H$ onto that coset, explaining the formula. If $|G|=p$ is prime, then $|H|$ is either $1$ or $p$, so $H=\{e\}$ or $H=G$. Thus the correct conclusion is no proper nontrivial subgroup. The trivial subgroup $\{e\}$ is proper for every group of prime order, so the wording in the question needs this qualification.

Let $d=\mathrm{ord}(g)$. The powersare distinct: equality of two with exponents between $0$ and $d-1$ would give a smaller positive power equal to $e$. Reducing exponents modulo $d$ proves closure and supplies inverses, so this is a cyclic subgroup of group order $d$. Lagrange's theorem now gives

Choose $g\ne e$. By Lagrange's theorem, its cyclic subgroup has order dividing the prime 101. It is not of order one, so $⟨g⟩=G$. Thus the group is cyclic, and hence abelian. In particular, every pair of elements is $g^i,g^j$, and $g^i{g}^j=g^{i+j}=g^{j+i}=g^j{g}^i$.

An algebraic element is a separable algebraic element over $K$ if its minimal polynomial of an algebraic element has distinct roots in a splitting field. A field extension is a separable extension if all its elements are separable algebraic elements.

The formal derivative is additive and satisfies $D(X^m)=m{X}^{m-1}$. For monomials, $D(X^r{X}^s)=(r+s)X^{r+s-1}=X^{r}D(X^s)+X^{s}D(X^r)$. Distributing over the finitely many coefficients proves the product rule $D(fg)=fDg+gDf$ over any field. If $f=(X-\alpha {)}^{2}h$ in an extension, the rule gives $Df=2(X-\alpha )h+(X-\alpha {)}^{2}Dh$, so $Df(\alpha )=0$, including in characteristic two. If $f,Df$ were relatively prime in $K[X]$, the Bézout's identity for polynomials would give $uf+vDf=1$, whose evaluation at $\alpha$ is impossible. Hence they are not relatively prime.

In characteristic zero, the formal derivative of a positive-degree irreducible polynomial $m$ is nonzero and has smaller degree. Thus $\gcd (m,Dm)=1$, since an irreducible polynomial has no proper nonconstant factor. By the previous argument, $m$ has no repeated root. Every element of a finite field extension is algebraic, so the extension is a separable extension.

For $K={\mathbb{F}}_p(t)$, let ${\alpha }^p=t$. The characteristic-$p$ identity gives $X^p-t=(X-\alpha {)}^p$, so $K(\alpha )$ is already a splitting field. It is a proper extension by the allowed no-root fact. More precisely, the minimal polynomial of an algebraic element $\alpha$ divides $X^p-t$ and, over $K(\alpha )$, equals $(X-\alpha {)}^r$ for some $1\le r\le p$. If $r<p$, its next-to-leading coefficient is $-r\alpha \in K$, forcing $\alpha \in K$ because $r\ne 0$ in ${\mathbb{F}}_p$, a contradiction. Hence its minimal polynomial of an algebraic element is $X^p-t$, with its one root repeated $p$ times. The extension is therefore inseparable, of degree $p$.

By contrast, every finite field extension $L/{\mathbb{F}}_p$ is a separable extension. If $[L:{\mathbb{F}}_p]=n$, then $|L|=p^n$. Lagrange's theorem in its nonzero multiplicative group gives $a^{p^n}=a$ for all $a\in L$. Thus every element is a root of $X^{p^n}-X$, whose formal derivative is $-1$; it has no repeated roots. Every minimal polynomial of an algebraic element divides this separable polynomial, proving the assertion.

Work in ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-d}]$, since $d\equiv 2(\mathrm{mod}4)$. First $x$ must be odd: if it were even, coprimality would make $a$ odd, so $x^2+a^{2}d\equiv 2(\mathrm{mod}4)$, which cannot be an odd power $y^p$ for $p\ge 3$. Thus $y$ is odd too. Also no odd prime $\ell$ can divide both $x$ and $d$. Such a prime does not divide $a$, and square-freeness of $d$ gives $v_{\ell }(x^2+a^{2}d)=1$, whereas $v_{\ell }(y^p)$ is divisible by $p$. Hence $\gcd (x,d)=1$.

Put $\alpha =x+a\sqrt{-d}$. Suppose a prime ideal $P$ divides both $(\alpha )$ and $(\overline{\alpha })$, and let $\ell$ be the rational prime below it. It then contains $2x$ and $2a\sqrt{-d}$. The case $\ell =2$ is impossible because $N((\alpha ))=y^p$ is odd. If $\ell$ is odd, the first inclusion gives $\ell \mid x$, and $\gcd (a,x)=1$ makes $a$ nonzero in the residue field. The second inclusion therefore gives $\sqrt{-d}\in P$, so $\ell \mid d$, contradicting the previous paragraph. Thus $(\alpha )$ and $(\overline{\alpha })$ are coprime ideals. In particular they generate the unit ideal, the precise meaning of coprimality here; no unique factorization of elements is assumed.

Their product is $(y{)}^p$. Unique prime ideal factorization and disjoint prime support imply $(\alpha )=A^p$ for an integral ideal $A$. In the ideal class group, $[A{]}^p=1$. Since $p$ does not divide the class number, Lagrange's theorem forces $[A]=1$. Write $A=(u+v\sqrt{-d})$. Equality of the principal ideals gives $\alpha =\epsilon (u+v\sqrt{-d}{)}^p$. The only units are $\pm 1$, since a unit satisfies $u^2+d{v}^2=1$ and $d\ge 2$. Both signs are $p$th powers for odd $p$, so they can be absorbed into the generator:Apply the class-number obstruction from a ramified odd exponent. For the equation with $a=1$ and $p\mid d$, compare the coefficients of $\sqrt{-d}$ in this identity. The coefficient on the right isIts $j=1$ term is divisible by $p$, and every $j\ge 3$ term contains a factor $d$, so the whole coefficient is divisible by $p$. The coefficient on the left is one, a contradiction. Thus there are no positive integer solutions under the stated class-number and divisibility assumptions.

Finally,The integer $174=2\cdot 3\cdot 29$ is square-free, is two modulo four, and is divisible by three. If its imaginary quadratic class number were not divisible by three, the preceding result with $p=3$ would prohibit this solution. Consequently

A monic quadratic polynomial over a field is irreducible exactly when it has no root in that field. Testing the nine choices of its two coefficients in ${\mathbb{F}}_3$ givesFor clarity, a zero constant term gives a root at zero. Among constant term one, only the zero linear coefficient avoids roots at $1,2$. Among constant term two, precisely the linear coefficients one and two avoid roots. This exhausts all choices.

Put $f=x^4+x+2$. Its values at $0,1,2$ are respectively $2,1,2$, so it has no linear factor. If this quartic were reducible, absence of a linear factor would force a factorization into two irreducible quadratics. Division by the three candidates above gives remainders respectively $x,x+1,x+1$, all nonzero. Thus $f$ is an irreducible polynomial.

The quotient by an irreducible polynomial is a field consequently gives a field of $3^4=81$ elements. Its nonzero elements form a multiplicative group of a finite field of order $80$. In this field, $a^4=1+2a$. Successive squaring givesThe latter is not one: otherwise the nonzero polynomial $1+x+2x^3$ of degree below four would vanish at $a$, contradicting its minimal polynomial. By Lagrange's theorem, $(a^{16}{)}^5=a^{80}=1$. Since five is prime, its order is exactly five. Hence one answer isIts other nonidentity powers are the remaining primitive fifth roots. It is unnecessary to prove that $a$ itself generates the entire multiplicative group.

The symmetric group $S_3$ has subgroups of orders $1,2,3,6$. There is one trivial subgroup, three conjugate subgroups generated by transpositions, one order-three subgroup $A_3$, and the whole group. These are all subgroups by Lagrange's theorem. Thus again there is one conjugacy class for each index $6,3,2,1$.

The coset action classification in part (c) supplies exactly one transitive orbit type of each of these sizes. The same orbit-multiplicity equation as in part (i) applies, givingThe equality of the two counts comes from their identical lists of transitive orbit types, rather than from the groups being isomorphic.