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Let $p\geq 3$ be prime and $n$ be a natural number. How can we prove the group of matrices

$$G = \left\{\left[\begin{array}{cc} a & b \\ 0 & d\end{array}\right] : a,b,d \in \mathbb{Z}/(p^n\mathbb{Z}),\quad ad = 1\right\}$$

has just one subgroup of order $p^{2n-1}?$

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The number of choices for $b$ is $p^n$.

The number of choice for $a$ is $\phi(p^n)=p^{n-1}(p-1)$.

Once $a$ is chosen, $d$ is uniquely determined.

It follows that $|G|=(p^n)\bigl(p^{n-1}(p-1)\bigr)=p^{2n-1}(p-1)$, hence $G$ has a $p$-Sylow subgroup of order $p^{2n-1}$.

Let $n_p$ be the number of distinct $p$-Sylow subgroups of $G$.

Then by Sylow theory, we get

  • $n_p{\,\mid\,}(p-1)$$\\[4pt]$
  • $n_p\equiv 1\;(\text{mod}\;p)$

hence $n_p=1$, which proves the claim.

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  • $\begingroup$ Thanks to your kind answer, I could understand the sentence after the fourth line . Thank you for the great answer ! But why can we calculate the number of choices a as Φ(p^n) ?Could you give me more explanation? $\endgroup$ – buoyant Aug 15 at 5:57
  • $\begingroup$ @buoyant: In order to have $ad=1$, $a$ must be a unit in $\mathbb{Z}/(p^n\mathbb{Z})$. Conversely, if $a$ is a unit in $\mathbb{Z}/(p^n\mathbb{Z})$, then $ad=1$ is achieved by letting $d=a^{-1}$. Thus, the number of choices for $a$ is equal to the number of units of $\mathbb{Z}/(p^n\mathbb{Z})$, which is $\phi(p^n)=p^{n-1}(p-1)$. $\endgroup$ – quasi Aug 15 at 6:04
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Hint: If you can guess what the subgroup is (I'll leave that as an exercise unless requested otherwise), you can easily check it is a normal $p$-Sylow subgroup. (Why does this suffice to show it's unique of its order?)

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