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.

  • 1
    $\begingroup$ 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 $4$. $\endgroup$
    – KCd
    Aug 10, 2021 at 2:47
  • $\begingroup$ @KCd Thanks for your comment. I was meant to let $p$ be an odd prime but somewhat forgot it… $\endgroup$ Aug 10, 2021 at 3:26

1 Answer 1


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 $M\phi(t)=\psi(t)M$.

  • $\begingroup$ Very nice argument! I really want to accept your answer now, but I want to wait for some other comments that tells me why my previous approach did not work. I will mark yours as accepted if no more threads are to be posted. $\endgroup$ Aug 10, 2021 at 3:32
  • $\begingroup$ 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 $\phi=\psi \circ f$. $\endgroup$ Aug 10, 2021 at 3:40
  • $\begingroup$ I see, thanks! The theorem I quoted is sufficient but not necessary. $\endgroup$ Aug 10, 2021 at 11:07

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