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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.

The polynomial $X^3+X+1$ has no root in ${\mathbb{F}}_2$, so it is an irreducible polynomial. The quotient ring of ${\mathbb{F}}_2[X]$ by the ideal it generates is therefore a field with eight elements. The seven nonzero elements form a group of order seven, so each satisfies $x^7=1$ by Lagrange's theorem. Thus this field splits $X^7-1$. A root different from one has order seven; it cannot lie in ${\mathbb{F}}_2$ and generates this degree-three field extension, since three is prime and the tower law applies. Therefore

The element $-1_R$ lies in the unit group and satisfies $(-1_R{)}^2=1_R$. Its order of a group element divides two and, by Lagrange's theorem, divides the odd number $|U(R)|$. Its order is therefore one, so $-1_R=1_R$ and $2\cdot 1_R=0$.

It follows thatis a well-defined unital ring homomorphism. For a nonzero ring, this says its characteristic of a ring is two. If the zero ring is permitted, the same map still exists. This is the odd unit group forces characteristic two argument, with that harmless exception to the characteristic wording.

The Sylow theorems also give $n_5\mid 14$ and $n_5\equiv 1(\mathrm{mod}5)$. Of $1,2,7,14$, only $1$ is congruent to one modulo five. Thus $H$ is a normal subgroup of $G$ as well.

By Lagrange's theorem, $H\cap K=\{1\}$, since its group order divides both $5$ and $7$. For $h\in H$ and $k\in K$, the group commutator $hk{h}^{-1}{k}^{-1}$ lies in $H$ by the normal subgroup property of $H$, and in $K$ by the normal subgroup property of $K$. It is therefore the identity element, so all elements of $H$ commute with all elements of $K$. This is the fact that normal subgroups of coprime order commute.

The subgroup $HK$ consequently has $|H||K|=35$ elements: the expression $hk$ is unique because $H\cap K=\{1\}$. Since $H$ and $K$ have prime orders, they are cyclic groups. Write $H=⟨h⟩$ and $K=⟨k⟩$. The product of commuting elements of coprime order has order $35$, so $hk$ generates $HK$. Finally, $HK$ is a normal subgroup because both its factors are normal subgroups; equivalently, it has subgroup index two. Therefore

Let $|G|=p^2$. The allowed center of a group theorem gives $|Z(G)|>1$. By Lagrange's theorem its order is $p$ or $p^2$. In the first case $G/Z(G)$ has order $p$ and is a cyclic group. A cyclic quotient by the center forces $G$ to be Abelian: if $G/Z(G)=⟨gZ(G)⟩$, write arbitrary elements as $g^{a}z,g^{b}w$ with $z,w\in Z(G)$, and commute them directly. Hence in every caseFor two examples at prime-cube order, take the dihedral group of a square, $⟨r,s\mid r^4=s^2=1,srs=r^{-1}⟩$, and the quaternion group $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$. Both have order $8=2^3$. They are nonabelian because $sr\ne rs$ and $ij=k\ne -k=ji$, respectively. They are distinct examples even up to group isomorphism: the dihedral group has five elements of order two and the quaternion group has only one. The request asks for two examples; it does not require the prime to be arbitrary.

Suppose $N$ were a normal subgroup of index $m$ and put $Q=SL(2,{\mathbb{F}}_p)/N$. The supplied reduction map and part (c) show that $Q$ is generated by the images $s,t$ of $S,ST$. Their orders divide $4$ and $6$, respectively, because $S^4=I$ and $(ST{)}^6=I$. By Lagrange's theorem, their orders also divide $|Q|=m$. Since $\gcd (m,6)=1$, both $\gcd (m,4)$ and $\gcd (m,6)$ equal one. Thus $s=t=1$, forcing $Q$ to be trivial and $m=1$, a contradiction. No such normal subgroup exists for $m>1$.

Let commuting elements $a,b$ have finite coprime orders $m,n$. Their cyclic subgroups have trivial intersection by Lagrange's theorem. If $(ab{)}^k=1$, then $a^k=b^{-k}$ belongs to this intersection, so $m\mid k$ and $n\mid k$. Thus $mn\mid k$. Conversely, $(ab{)}^{mn}=1$ because the elements commute. This proves the displayed order formula.

A finite nonabelian simple group embeds in $A_{n_p}$, where $n_p$ is its number of Sylow subgroups for a prime dividing its order. It is not a p-group by the nontrivial-centre property. Thus its Sylow subgroups are proper and cannot be normal, so $n_p>1$. The conjugation action is nontrivial, and its normal kernel of a group homomorphism must therefore be trivial. The sign of this faithful permutation action must be trivial, since a nontrivial sign homomorphism would inject the group into a group of order two. Lagrange's theorem consequently gives $|G|\mid n_p!/2$.

With the generators $c,a$ of the direct product of cyclic groups of orders four and two, the subgroups are the trivial group, the whole group, the three order-two groups $⟨c^2⟩$, $⟨a⟩$, $⟨c^{2}a⟩$, and the three order-four groups $⟨c⟩$, $⟨ca⟩$, $⟨c^2,a⟩$. These are exhaustive by Lagrange's theorem: a proper nontrivial subgroup has order two or four; an order-four subgroup either contains an element of order four and is one of the two cyclic examples, or is precisely the four-element set killed by squaring. The latter is a Klein four-group.