I'm stuck in the classification of groups of order $p^3$. These were the steps i followed.
I showed that if $|Z(G)|=p^2$ or $p^3$ then G is abelian. Hence order of centre is $p$. As centre is normal, G can be written as a semi direct product of $H\times K$($\times$ is semidirect product symbol) where $H$ is $Z(G)$ and $K$ is a group of order $p^2$. And let π be the homomorphism from K to Aut(H). Aut(H) is isomorphic to $Z_{p-1}$. Now there are 2 cases depending on what K is:

(1) K is $Z_{p^2}$ Now π : $Z_{p^2 }$ --> $Z_{p-1}$ There aren't any homomorphisms(except the trivial) possible as no element in $Z_{p-1}$ has order $p$ or $p^2$.

(2) K is $Z_p\times Z_p$ Again the same argument as in (1) will hold and no homomorphisms will be there except the trivial. If it is trivial homomorphism then group becomes abelian($Z_{p^2}\times Z_p $ and $Z_p\times Z_p \times Z_p $ respectively) I ended up proving that there is no nonabelian group of order $p^3$ which is not true, hence the proof is wrong somewhere.I couldn't find the mistake. Any help/partial progress will be appreciated.


Since $H$ is the center, $H$ commutes with everything so with $K$ also. Thus the resulting homomorphism is trivial.

Moreover, there is no $K$ in $F$ such that $G=H\times K$ since every normal subgroup of $G$ intersects nontrivially with $Z(G)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.