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

by buoyant   Last Updated August 14, 2019 09:20 AM - source

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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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?)

runway44
August 14, 2019 08:53 AM

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.

quasi
August 14, 2019 08:56 AM