If $G=\langle a, b : a^{p^{n}}= 1= b^{p^{n+1}}, [b, a]= b^{p} \rangle$, such that $n\geq 2$ and $p$ is an odd prime number, then how can I define a non-inner automorphism of $G$? Is it possible to find the structure of all automorphisms of $G$?
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1$\begingroup$ Which convention do you use for $[b,a]$? $\endgroup$– YCorCommented Jul 17, 2017 at 20:54
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2$\begingroup$ $a \mapsto a$, $b \mapsto b^{-1}$ defines an automorphism of order $2$, which must be non-inner. Computer calculations indicate that the automorphism group has order $(p-1)p^{2n+1}$, so the outer automorphism group has order $(p-1)p$. The subgroup of order $p-1$ is coming from automorphisms that fix $a$ and map $b$ to a power of itself. $\endgroup$– Derek HoltCommented Jul 17, 2017 at 21:49
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$\begingroup$ OK, there are indeed 4 conventions for $[x,y]$, namely (writing $x'=x^{-1}$) $xyx'y'$, $yxy'x'$, $x'y'xy$, $y'x'yx$. For these conventions, the relation $[b,a]=b^p$ reads as resp. $aba'=b^{p+1}$, $aba'=b^{1-p}$, $a'ba=b^{1-p}$, $a'ba=b^{1+p}$. Since the order of both $1+p$ and $1-p$ modulo $p^{n+1}$ is exactly $p^n$ for odd $p$, all conventions yield the same group, namely the semidirect product $(\mathbf{Z}/p^{n+1}\mathbf{Z})\rtimes C$, where $C$ is the subgroup (of index $p-1$; cyclic of order $p^n$) in $(\mathbf{Z}/p^{n+1}\mathbf{Z})^\times$ consisting of those elements $\equiv 1$ mod $p$. $\endgroup$– YCorCommented Jul 18, 2017 at 13:39
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I'm assuming $[x,y]=x^{-1}y^{-1}xy$, but I expect that using the other convention you would get similar results.
Note that $a^{-1}ba = b^{1+p}$, and hence $a^{-1}b^pa = b^{p+p^2}\in\langle b^p\rangle$. Thus, $\langle b^p\rangle\triangleleft G$, and $G/\langle b^p\rangle$ is abelian, so $G$ is $2$-generated and has cyclic commutator subgroup.
For odd prime, such groups were studied by Miech:
Miech R.J. On $\mathbf{p}$-groups with a cyclic commutator subgroup. J. Austral. Math. Soc. 20 (1975), no. 2, 178–198. MR0404441 (53 #8243)
Miech describes every such group in terms of a 12-tuple of integers satisfying certain conditions. Namely:
Theorem 1. Let $p$ be an odd prime and let $G$ be a finite nonabelian $p$-group generated by two elements such that the commutator subgroup of $G$ is cyclic. Then $G$ has a pair of generators $\{x,y\}$ such that the defining relations of the group are: $$\begin{alignat*}{3} [y,x]&= z, &\quad x^{p^{\alpha}} &= z^{Rp^{\rho}}, &\quad y^{p^{\beta}} &= z^{Sp^{\sigma}} \\ z^{p^{\gamma}} &= 1, &\quad [z,x] &= z^{Mp^{\mu}}, &\quad [z,y] &= z^{Np^{\nu}} \end{alignat*}$$ where $\alpha,\beta,\gamma,\rho,\sigma,\mu,\nu$ and $R,S,M,N$ are integers that satisfy the conditions $$ \alpha\geq \beta, \quad \alpha\geq \gamma,\quad \rho+\mu\geq \gamma,\quad \rho+\nu\geq \gamma,\quad \sigma+\nu\geq \gamma,\\ 1\leq \mu,\quad \nu\leq\gamma,\quad 0\leq \rho,\quad \sigma\leq\gamma,\quad p^{\beta} - MSp^{\mu+\sigma}\equiv 0\pmod{p^{\gamma}}.$$ Conversely, given any set of parameters $\{\alpha,\beta,\gamma,\rho,\sigma,\mu,\nu,R,S,M,N\}$ that satisfies these conditions there is a group defined by these relations.
Your groups would then be given by $\alpha=n$, $\beta=1$, $\gamma=n$, $\rho=n$, $\sigma=0$, $\mu=1$, $\nu=n$, and $R=S=M=N=1$.
If you take $\alpha,\beta,\gamma$ as fixed, the groups can be succinctly described then by a $4$-tuple, $[Rp^{\rho}, Sp^{\sigma}, Mp^{\mu}, Np^{\nu}]$.
Section 1 of the paper describes the isomorphisms/automorphisms in Theorem 9 in terms of four congruences (a bit hard to type them here, since I've had to change some of the notation to avoid using $a$ and $b$, which you use in your description). This should allow you to count the automorphisms and figure out which ones are given by conjugation, though it will probably take some number crunching.
answered Jul 18, 2017 at 4:49