Uniqueness of Semidirect Product $(\mathbb{F}_p\times\mathbb{F}_p)\rtimes\mathbb{F}_p$

Let $$p$$ be an odd prime. For a non-abelian group $$G$$ of order $$p^3$$ in which no element is of order $$p^2$$, it is isomorphic to a semidirect product $$(\mathbb{F}_p\times\mathbb{F}_p)\rtimes\mathbb{F}_p.$$ Here I am trying to see if such semidirect product is unique up to isomorphism.

Note that $$\operatorname{Aut}(\mathbb{F}_p\times\mathbb{F}_p)\cong\operatorname{GL}(2,\mathbb{F}_p),$$ so for convenience, we shall consider a nontrivial group homomorphism $$\phi:\mathbb{F}_p\to\operatorname{GL}(2,\mathbb{F}_p)$$.

Since $$\phi$$ is nontrivial, it must be injective in this case, so $$\phi(1)=:A$$ is of order $$p$$. It has been shown from this post that $$A$$ has Jordan canonical form $$J:=J_2(1)=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$. That is, $$A=BJB^{-1}$$ for some $$B\in\operatorname{GL}(2,\mathbb{F}_p)$$.

Now let $$\psi:\mathbb{F}_p\to\operatorname{GL}(2,\mathbb{F}_p)$$ be another nontrivial group homomorphism. Then $$\psi(1)=CJC^{-1}$$ for some $$C\in\operatorname{GL}(2,\mathbb{F}_p)$$. By the following theorem:

Theorem. Let $$H$$ and $$K$$ be two groups and $$\phi,\psi:K\to\operatorname{Aut}(H)$$ be group homomorphisms. If $$\psi=\phi\circ f$$ for some $$f\in\operatorname{Aut}(K)$$, then $$H\rtimes_\phi K\cong H\rtimes_\psi K.$$

it suffices to find some $$f\in\operatorname{Aut}(\mathbb{F}_p)$$ such that $$\psi=\phi\circ f$$. Suppose $$f(1)=k$$. Then $$(\phi\circ f)(1)=\phi(f(1))=\phi(k)=\phi(1)^k=(BJB^{-1})^k=BJ^kB^{-1}.$$ If $$\psi=\phi\circ f$$, then we shall have $$CJC^{-1}=\psi(1)=(\phi\circ f)(1)=BJ^kB^{-1}\implies (B^{-1}C)J=J^k(B^{-1}C).$$ Let $$S:=B^{-1}C=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$$. It follows that $$\begin{bmatrix} a & a+b \\ c & c+d \end{bmatrix}=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}=SJ=J^kS=\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix}=\begin{bmatrix} a+kc & b+kd \\ c & d \end{bmatrix}.$$ Then we have $$c=0$$ and $$a=kd$$, which are not necessarily satisfied since $$B$$ and $$C$$ are arbitrary. Then I got stuck with this step.

I'd appreciate it if anyone has good ideas on this.

• When $p = 2$ there is no non-abelian group of order 8 in which no element has order $p^2$: the quaternion and dihedral groups of order $8$ both have elements of order $4$.
– KCd
Aug 10, 2021 at 2:47
• @KCd Thanks for your comment. I was meant to let $p$ be an odd prime but somewhat forgot it… Aug 10, 2021 at 3:26

So far, you have proven that $$\phi$$ and $$\psi$$ differ by $$\phi(t)=M^{-1} \psi(t) M$$, where $$M \in \operatorname{Aut}(\mathbb{F}^p \times \mathbb{F}^p)$$. (Here $$M=CB^{-1}$$ if I'm not mistaken, but we will not need the explicit expression.) Now, you can use this to construct a explicit isomorphism. I imagine that this is similar in spirit to the theorem you quoted anyway.
Consider $$\Phi\colon H \rtimes_\phi K \to H \rtimes_\psi K, \qquad \Phi(h, k)=(Mh, k).$$ This is of course bijective, so we only have to check that the map is a group homomorphism. In fact, we have $$\Phi(h_1, k_1) \cdot \Phi(h_2, k_2)=(Mh_1, k_1) \cdot (Mh_2, k_2)=(Mh_1 \cdot \psi(k_1)(Mh_2), k_1\cdot k_2),$$ and $$\Phi((h_1, k_1) \cdot (h_2, k_2))= \Psi(h_1 \cdot\phi(k_1)(h_2), k_1\cdot k_2)= (Mh_1 \cdot M \phi(k_1)(h_2), k_1 \cdot k_2).$$ But these expressions are equal, as $$M\phi(t)=\psi(t)M$$.
• I think that the problem with your approach was that, if $\phi=\psi \circ f$, then in particular $\phi$ and $\psi$ have the same image (as $f$ is bijective). But this is not necessarily true: for instance, the transpose of $J$ also has order $p$, but is not contained in $\{J, J^2, J^3, \dots\}$. Thus, the maps $\phi\colon 1 \mapsto J$ and $\psi\colon 1 \mapsto J^T$ cannot satisfy the condition $\phi=\psi \circ f$. Aug 10, 2021 at 3:40