Let $p$ and $q$ be twin primes, both greater than or equal to $5$ , is every group of order $p^2q^2$ abelian?

Question

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?

Answer 1

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,

• since $B$ is normal, then $ab \in aB = Ba$ and $ab = b'a$ for some $b' \in B$, and
• since $A$ is normal, then $ab \in Ab = bA$ and $ab = ba'$ for some $a' \in A$.

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.

Answer 2

With a more general spin: yes, for twin primes $p$ and $p+2$, where $p \geq 5$, all groups of order $p^2(p+2)^2$ are abelian indeed! In general, one can show that every group of order $n$ is abelian, if and only if $n$ satisfies the following criterion: $n=p_1^{e_1}\cdots p_r^{e_r}$ with distinct primes $p_i$, $e_i \in \{1,2\}$ (so $n$ must be cubefree), and $p_i$ does not divide $p_j^{e_j}-1$ for all $i$ and all $j$. This is a theorem of L.E. Dickson from $1905$.

In the situation above, one has to check two cases: $p^2 \not \equiv 1$ mod $p+2$, and $(p+2)^2 \not \equiv 1$ mod $p$. I leave the first to you to prove. The last, if we would have $(p+2)^2 \equiv 1$ mod $p$, then $2^2 \equiv 1$ mod $p$, leading to $p \mid 3$, that is $p=3$, a contradiction with $p \geq 5$.

Indeed $225=3^25^2$, and $3\mid 5^2-1$, so not every group of order $225$ is abelian.

For more information on abelian, and also nilpotent and cyclic numbers $n$, see this paper and also this post on Math StackExchange.