Past exam of the short options course of the University of Oxford / 2017 / 2

Use the fact that $K$ is a normal subgroup to move a factor of $K$ past a factor of $H$. For $h_1,h_2\in H$ and $k_1,k_2\in K$,whileThe identity element belongs to $HK$. Hence $HK$ is a subgroup, by the subgroup test. This proves the product of a subgroup with a normal subgroup construction; $H$ need not be a normal subgroup.

For each $h\in H$, conjugation maps $K$ to itself and respects its multiplication. The inverse of $k\mapsto h^{-1}kh$ is $k\mapsto hk{h}^{-1}$, so both are group automorphisms of $K$.

The printed formula requires a composition convention. For the printed ${\varphi }_h(k)=h^{-1}kh$, ordinary function composition givesIndeed, $(h_1{h}_2{)}^{-1}k(h_1{h}_2)=h_2^{-1}(h_1^{-1}k{h}_1)h_2$. Thus the displayed assignment is a group antihomomorphism with the usual multiplication in the automorphism group. It is a group homomorphism if the automorphisms are written as right operators and multiplied in the corresponding reverse order; this is precisely why right conjugation reverses composition.

With the usual left function composition, the corresponding group homomorphism is insteadDirect expansion gives ${\psi }_{h_1{h}_2}(k)=h_1(h_{2}k{h}_2^{-1})h_1^{-1}=({\psi }_{h_1}\circ {\psi }_{h_2})(k)$. This states the intended conjugation action without silently reversing the order in the printed formula.

Write the group order as $|G|=p^{a}m$, where $p$ is a prime number and $p\nmid m$. A Sylow p-subgroup is a subgroup of group order $p^a$, the largest power of $p$ dividing $|G|$.

The three Sylow theorems assert:

In particular, a unique Sylow p-subgroup is a normal subgroup, since all its conjugates must equal it.

The Sylow theorems give $n_7\mid 10$ and $n_7\equiv 1(\mathrm{mod}7)$. Among the divisors $1,2,5,10$, only $1$ satisfies the modular congruence. HenceUniqueness and the conjugacy assertion of the Sylow theorems make $K$ a normal subgroup.

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

Put $N=HK=⟨r⟩$, the normal subgroup of order $35$. By Cauchy's theorem for finite groups, $G$ contains an element $t$ of order two. Such an element cannot belong to the odd-order subgroup $N$. Thus $N\cap ⟨t⟩=\{1\}$ and $G=N⟨t⟩$, giving the semidirect productEvery automorphism of a finite cyclic group sends $r$ to $r^a$ for a unit $a$ modulo $35$. The relation $t^2=1$ forces the conjugation action of $t$ to square to the identity:The Chinese remainder theorem identifies a congruence class modulo $35$ with its pair of classes modulo $5$ and $7$. In each of these prime fields, the factorization $(a-1)(a+1)=0$ forces $a=1$ or $a=-1$. The choices are independent, giving exactly four homomorphisms from a finite cyclic group $C_2\to \mathrm{Aut}(C_{35})$. Explicitly,The factors $C_5$ and $C_7$ of $N$ are preserved by every automorphism, since each is its unique subgroup of that order. The four conjugation actions therefore give:

Here $D_n$ has group order $n$, as in the paper. The center of a group distinguishes these four groups of order seventy: its orders are $70,1,5,7$, respectively, so the four groups are pairwise nonisomorphic. Each construction is a semidirect product with $70$ elements, so every listed possibility occurs.