In a classification of groups of order $p^3q$, where $p, q$ are distinct primes, when considering non-nilpotent groups with normal Sylow $p$-subgroups they mention that these groups have the form $N \rtimes U$, with $|N|=p^3$ and $U=C_q$. Then there is a case where,
1) $N$ is extraspecial of exponent $p$: Then $Aut(N) \rightarrow Aut(N/ \phi(N)) \cong GL(2,p)$ is surjective. Thus the conjugacy classes of subgroups of order $q$ in $Aut(N)$ correspond to the conjugacy classes of subgroups of order $q$ in $GL(2,p)$.
and,
2) $N$ is extraspecial of exponent $p^2$: Then $Aut(N)$ is solvable and has a normal Sylow $p$-subgroup with $p$-complement of the form $C_{p-1}$. Thus there is atmost one subgroup of order $q$ in $Aut(N)$ and this exists iff $q|(p-1)$.
Can someone please help me to identify how to write the above two groups in notation form. i.e. Can the groups be written as follows using notations:
1) $N \cong C_{p^2} \times C_p$
2) $N \cong C_{p^3}$
What is meant by "extraspecial of exponent". Please help me with this question.
Thanks a lot in advance.
An extraspecial $p$-group is a nonabelian group $N$ such that the center $Z(N)$ is cyclic of order $p$ and $N/Z(N)$ is an elementary abelian $p$-group, i.e. it is isomorphic to $C_p^n$ for some $n$. So, an extraspecial group of exponent $p$ is an extraspecial $p$-group whose exponent is $p$, and an extraspecial group of exponent $p^2$ is an extraspecial $p$-group whose exponent is $p^2$. (These are the only possible exponents for an extraspecial $p$-group, since $Z(N)$ and $N/Z(N)$ both have exponent $p$ and so $x^{p^2}=1$ for all $x\in N$.)
The extraspecial groups can be completely classified; you can find more information at the link above. In particular, for $p$ odd, an extraspecial group of order $p^3$ and exponent $p$ is isomorphic to the group of matrices of the form $\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1\end{pmatrix}$ over $\mathbb{F}_p$, and an extraspecial group of order $p^3$ and exponent $p^2$ is a semidirect product $C_{p^2}\rtimes C_p$ where the action of $C_p$ on $C_{p^2}$ is nontrivial.