Question

If $\mathbf{\left|}\mathbf{G}\mathbf{\right|}\mathbf{=}{\mathbf{p}}^{\mathbf{n}}$, prove that every subgroup of G of order ${\mathbf{p}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$ is normal.

Short answer

It is proved that, every subgroup of G of order ${\mathrm{p}}^{\mathrm{n}-1}$ is normal.

Step-by-step solution

Apply Mathematical Induction

We prove this by induction on n .

If n=1, then $G={\mathrm{\mathbb{Z}}}_p$; the only subgroup of order $p^{n-1}=1$is $\left\{e\right\}$and it is normal.

Suppose $n>1$and that every subgroup of order ${\mathrm{p}}^{\mathrm{k}-1}$of a group of order $p^k$is normal for all .

Let H be a subgroup of order $p^{n-1}$of G.

Apply Theorem 9.27

Theorem 9.27 says that if G is a group of order pⁿ with p prime and $\mathit{n}\mathbf{\ge }\mathbf{1}$, then the center $\mathbf{Z}\left(G\right)$ contains more than one element. In particular, $\left|Z\left(G\right)\right|\mathbf{=}{\mathbf{p}}^{\mathbf{k}}$with $\mathbf{1}\mathbf{\le }\mathbf{k}\mathbf{\le }\mathbf{n}$.

By Theorem 9.27, the center $Z\left(G\right)$of G is a non-trivial normal subgroup of G.

If $Z\left(G\right)\not\subset H$then, $\mathrm{HZ}\left(\mathrm{G}\right)=\mathrm{G}$.

So, every $\mathrm{x}\in \mathrm{G}$can be written as $\mathrm{x}=\mathrm{hz}$for $\mathrm{h}\in \mathrm{H}$and role="math" localid="1659427768018" $z\in \mathrm{Z}\left(\mathrm{G}\right)$.

So that, ${\mathrm{x}}^{-1}\mathrm{Hx}={\mathrm{z}}^{-1}{\mathrm{h}}^{-1}\mathrm{Hhz}=\mathrm{H}$implies H is normal in G.

If $Z\left(G\right)\subseteq H$then we have that $G|Z\left(G\right)$ is a group of order p^k for some data-custom-editor="chemistry" $\mathrm{k}<\mathrm{n}$and data-custom-editor="chemistry" $\mathrm{H}|\mathrm{Z}\left(\mathrm{G}\right)$is a subgroup of order data-custom-editor="chemistry" ${\mathrm{p}}^{\mathrm{k}-1}$.

By the inductive hypothesis, $H|Z\left(G\right)$is normal in $G|Z\left(G\right)$.

From this, we can conclude that for all $x\in G$and $h\in H$, there is some $z\in Z\left(G\right)$such that ${\mathrm{x}}^{-1}\mathrm{hx}=\mathrm{hz}\in \mathrm{h}$and therefore, ${\mathrm{x}}^{-1}\mathrm{HX}=\mathrm{H}$.

Since $x\in G$was arbitrary, we conclude that H is normal in G.

Thus, every subgroup of G of order $p^{n-1}$ is normal.