$G$ 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 $p^{2n-1}?$

Tags : group-theory

Answers 2

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?)

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.

August 14, 2019 08:56 AM

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