I don't know if I made a mistake, but it seems that something is missing here.
Let $p$ and $q$ be twin primes, i.e., $q = p + 2$. We want to prove that every group with order $p^2q^2$ is abelian, where $p$ and $q$ are greater than or equal to 5.
Call our group $G$ with $|G| = p^2q^2$. Let $n_p$ be the number of subgroups of $G$ with order $p^2$, and let $n_q$ be the number of subgroups of $G$ with order $q^2$.
We have that $n_p$ divides $q^2$ and $n_p$ is congruent to 1 mod $p$. This gives us initially $n_p = (1, q, q^2)$.
We have that $n_q$ divides $p^2$ and $n_q$ is congruent to 1 mod $q$. This gives us initially $n_q = (1, p, p^2)$.
But since $q = p + 2$:
If $n_p = q = p + 2$, then $p + 2 – 1$ is not congruent to 0 mod $p$.
If $n_p = q^2 = (p + 2)^2 = p^2 + 4p + 4$, then $p^2 + 4p + 4 – 1$ is not congruent to 0 mod $p$ because $p > 3$.
So, $n_p = 1$. Let's call $A$ the subgroup of $G$ with order $p^2$.
Also, if $q = p + 2$, then $p = q – 2$:
If $n_q = p = q – 2$, then $q – 2 – 1$ is not congruent to 0 mod $q$, because $q > 3$.
If $n_q = p^2 = q^2 – 4q + 4$, then $q^2 – 4q + 4 – 1$ is not congruent to 0 mod $q$ because $q > 3$.
So, $n_q = 1$. Let's call $B$ the subgroup of $G$ with order $q^2$.
A $p$-Sylow subgroup is normal in $G$ if and only if $n_p = 1$. Thus, $A$ and $B$ are normal in $G$.
Furthermore, $\text{gcd}(p^2, q^2) = 1$, which implies that $A \cap B = \{1\}$.
This gives us $|AB| = |A \cdot B| = p^2q^2 = |G|$.
Since $\text{gcd}(p^2, q^2) = 1$, by the Chinese remainder theorem, $Z_{p^2} \times Z_{q^2}$ is isomorphic to $Z_{p^2q^2}$.
Groups of order $p^2$ with $p$ prime are abelian, so $Z_{p^2}$ and $Z_{q^2}$ are abelian. The direct product of abelian groups is abelian. Thus, $Z_{p^2q^2}$ is abelian.
Am I missing something?
Best Answer
Your argument has some errors. In particular you claim to have proved that every group of order $p^2q^2$ is isomorphic to $\mathbb Z_{p^2q^2}$, but there are other 3 non-ismorphic abelian groups of order $p^2 q^2$, namely: $\mathbb Z_{p^2q} \times \mathbb Z_q$, $\mathbb Z_{pq^2} \times \mathbb Z_p$, and $\mathbb Z_{pq} \times \mathbb Z_{pq}$.
Your proof goes correctly until the sentece: "Since ${\rm gcd}(p^2,q^2) = 1, \ldots$". Until there you have correctly proved that both $A$ and $B$ are normal subgroups of orders $p^2$ and $q^2$, and, hence, both $A$ and $B$ are abelian. Moreover, you have obtained that the Frobenius product $A \cdot B = G$ and that $A \cap B = \{1\}$.
Now, you should prove that $A \cdot B$ is abelian. Since both $A$ and $B$ are abelian, it only remains to check that $ab = ba$ for all $a \in A$ and all $b \in B$. This follows from the already proven facts: (1) both $A$ and $B$ are normal subgroups of $G$, (2) $A \cap B = \{1\}$. Indeed,
Thus, $ab = ba' = b'a$ and we get $(b')^{-1}b = a (a')^{-1}$ and, using that $A \cap B = \{1\}$, we conclude that $a = a',\, b = b'$ and $ab = ba$. This finishes the proof.