2. 1. Show that a group whose order is the square of a prime is Abelian. [You may assume that a group of prime power order has a nontrivial centre.] Give two examples of non-Abelian groups whose order is the cube of a prime.
   2. Let $G$ be a group whose order $|G|$ is the product of two distinct primes. Stating carefully any theorems that you need, show that $G$ has a nontrivial cyclic normal subgroup. What does this tell you about the structure of $G$ ?
   3. Let $|G|=43681=11^2 19^2$. Show that $G$ is Abelian and list all the possible groups $G$.
   4. Now let $|G|=11^2 p^2$ where $p ≠ 11$ is prime. For which $p$ can we still conclude that $G$ must be Abelian?
   5. Give an example, with justification, of an odd prime $p$ such that there is a non-Abelian group of order $11^2 p^2$.

Solution

2. 1. First suppose that $G$ contains an element $x$ of order $p^2$. Then ${|⟨x⟩|}={|G|}$, and hence $G≅C_{p^2}$.
      Now suppose that $G$ does not contain an element of order $p^2$. Let $x$ be any element in $Z(G)∖\{1\}$. Then ${|x|}={|⟨x⟩|}=p$. Pick any element $y ∈ G∖⟨x⟩$. Then again the subgroup $⟨y⟩$ has order $p$, and Lagrange guarantees that $⟨x⟩∩⟨y⟩=\{1\}$. Moreover, $⟨x, y⟩$ is a subgroup of order $p^2$ of $G$ (again Lagrange) and hence coincides with $G$. Since $x∈Z(G)$, the following map is an isomorphism: $$ ϕ:⟨x⟩×⟨y⟩→G,  (x^a, y^b) ↦ x^a y^b $$Thus, $G ≅⟨x⟩×⟨y⟩≅ C_p×C_p$ in this case.
   2. Let $p>q$ be primes. $n_p≡1\pmod p,n_p|q⇒n_p=1$. So there is a unique Sylow $p$-subgroup, which is normal and isomorphic to $C_p$.

   3. Sylow's theorem	p=11	p=19
      1	$P_{11}=C_{11}×C_{11}$ or $C_{11^2}$	$P_{19}=C_{19}×C_{19}$ or $C_{19^2}$
      2	all conjugate
      3	$n_{11}≡1\pmod{11},n_{11}|19^2$ $⇒n_{11}=1$	$n_{19}≡1\pmod{19},n_{19}|11^2$ $⇒n_{19}=1$

      both normal$⇒G=P_{11}×P_{19}$
   4. $\left.\begin{array}rn_{11}≠1\\n_{11}|p^2\\n_{11}≡1\pmod{11}\end{array}\right\}⇔p≡±1\pmod{11}$
      $\left.\begin{array}rn_p≠1\\n_p|11^2\\n_p≡1\pmod p\end{array}\right\}⇔11≡±1\pmod p$
      Altogether $n_p=n_{11}=1$ if $p≠2,3,5$ and $p≢±1\pmod{11}$
   5. Simplify: look for $|H|=11p$. By (b), $H=C_{11}⋊C_p$. This is non-Abelian if $C_p$ acts nontrivially. We need $p|\operatorname{Aut}C_{11}≅(ℤ/11ℤ)^×=10$. But $p$ is odd prime, take $p=5$.