# $G$ has only one subgroup of order $p^{2n-1}$

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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## 2 Answers

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.

• 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? – buoyant Aug 15 at 5:57
• @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 $\phi(p^n)=p^{n-1}(p-1)$. – quasi Aug 15 at 6:04

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?)