How to prove this element is central Let $G=N\rtimes H$, where $$H=\langle b\rangle\cong C_{p^2}$$ and $$N= \langle a_1\rangle\times \langle a_2\rangle\times… \times\langle a_{p-1}\rangle\cong C_{p^2}\times C_p\times …\times C_p.$$
The action is given by
$$a_1^b=a_1a_2, a_2^b=a_2a_3, … , a_{p-1}^b=a_{p-1}a_1^{-p}.$$
How can I prove that $b^p$ is central? If $b^p$ commutes with every $a_i$, then it is central. I tried computing $a_1^{b^p}$ in the following way $a_1^{b^p}= (a_1^b)^{b^{p-1}}=(a_1a_2) ^{b^{p-1}}= a_1 ^{b^{p-1}} a_2^{b^{p-1}} =…$. But there was too many calculations. Is there another way?
The commutator subgroup is $G’=\langle a_2,…,a_{p-1},a_1^{-p}\rangle$. And for every element $g\in G$, the $p$-th power of $g$ can be written in the form $g^p=(b^ia_1^j)^p$.
 A: I prefer to work additively instead of multiplicatively.
The abelian group
$$
  ℤ/p^2 ⊕ \underbrace{ ℤ/p ⊕ \dotsb ⊕ ℤ/p }_{p - 2}
$$
is generated by the standard basis vectors $e_1, \dotsc, e_{p - 1}$ (corresponding to $a_1, \dotsc, a_{p - 1}$ in your notation).
We need to show that the endomorphism $φ$ given by
$$
  φ(e_1) = e_1 + e_2 \,,
  \quad
  \dotsc \,,
  \quad
  φ(e_{p - 2}) = e_{p - 2} + e_{p - 1} \,,
  \quad
  φ(e_{p - 1}) = e_{p - 1} - p e_1
$$
satisfies $φ^p = \mathrm{id}$.
We will show that $φ^p(e_n) = e_n$ for every $n = 1, \dotsc, p - 1$.
We observe that
\begin{align*}
  φ(e_1)   &= e_1 + e_2  \,, \\
  φ^2(e_1) &= e_1 + 2 e_2 + e_3 \,, \\
  φ^3(e_1) &= e_1 + 3 e_2 + 3 e_3 + e_4 \,, \\
  &\enspace\vdots
\end{align*}
The occurring coefficients are precise the binomial coefficients.
This is not just a coincidence:
if
$$
  φ^q(e_n) = {} \dotsb + a_{k-1} e_{k-1} + a_k e_k + \dotsb {} \,,
$$
then it follows with $φ(e_{k-1}) = e_{k-1} + e_k$ and $φ(e_k) = e_k + e_{k + 1}$ that the coefficient of $e_k$ in $φ^{q + 1}(e_n)$ will be $a_{k - 1} + a_k$.
The iterated application of $φ$ to a basis vector $e_n$ can thus be computed as
$$
  φ^q( e_n ) = \sum_{k = 0}^q \binom{n}{k} e_{n + k}
  \qquad
  \text{for $q < n - p$} \,.
$$
For larger values of $q$ we run into problems because we would require $φ(e_{p - 1}) = e_{p - 1} + e_p$, but instead have $φ(e_{p - 1}) = e_{p - 1} - p e_1$.
However, through the magic of notation, we can extend the above formula for $φ^q(e_n)$ to arbitrary large exponents $q$.
Let us set
$$
  e_p ≔ - p e_1 \,,
$$
so that $φ(e_{p - 1}) = e_{p - 1} + e_p$.
We observe that
$$
  φ(e_p) = φ(-pe_1) = -p φ(e_1) = -p e_1 - p e_2 = -p e_1 = e_p \,,
$$
which means that $e_p$ is fixed by $φ$.
We can therefore set
$$
  e_n ≔ 0
  \qquad
  \text{for every $n > p$}
$$
to get that
$$
  φ(e_n) = e_n + e_{n + 1}
  \qquad
  \text{for every $n ≥ 1$} \,.
$$
We can now state without restriction that
$$
  φ^q( e_n ) = \sum_{k = 0}^q \binom{n}{k} e_{n + k}
$$
for every $n ≥ 1$ and every exponent $q ≥ 1$.
Let us now consider the case $q = p$.
We find that
$$
  φ^p(e_n)
  =
  e_n
  + \binom{p}{1} e_{n + 1}
  + \binom{p}{2} e_{n + 2}
  + \dotsb
  + \binom{p}{p - 1} e_{n + p - 1}
  + e_{n + p}
$$
for every $n ≥ 1$.
The coefficient $\binom{p}{k} = p! / k! (n - k)!$ is divisible by $p$ for every $k = 1, \dotsc, p - 1$ because $p$ is prime (the numerator $p!$ is divisible by $p$, but the denominator $k! (n - k)!$ is not).
We observe that $p e_m = 0$ for every $m ≥ 2$, since $e_m$ is either contained in $ℤ/p$, given by the element $p e_1$ of $ℤ/p^2$, or zero to begin with.
We hence find that
$$
  φ^p(e_n) = e_n + e_{n + p} \,.
$$
But we also have $e_{n + p} = 0$ because $n + p > p$.
Therefore,
$$
  φ^p(e_n) = e_n \,.
$$
