I'm stuck on this exercise from Herstein's topics in algebra:

Suppose that $G$ is a group and $|G| = pm$, where $p \not| ~ m$ and $p$ is a prime. If $H$ is a normal subgroup of order $p$ in $G$, prove that $H$ is characteristic.

How to prove that? What we know:

  • $H$ and $\varphi(H)$ have prime order $p$, so they must be cyclic and abelian
  • They are both normal subgroups of $G$

how to show that, in fact, we have $H = \varphi(H)$ for every automorphism $\varphi$ ?

  • 1
    $\begingroup$ Do you know about Sylow subgroups yet? Subgroups of order $p$ are Sylow $p$-subgroups in $G$, so $\varphi(H)$ must be conjugate to $H$. But $H$ is normal, so... $\endgroup$ – Arturo Magidin Feb 22 '12 at 18:32
  • $\begingroup$ Hi, the question is quite early in the book and the few things I know are: Definitions and Examples of Groups \ Subgroups \ Lagrange's Theorem \ Homomorphisms and Normal Subgroups. I understand that it might be difficult to prove that using only basic facts.. Herstein seems to love this kind of question $\endgroup$ – Haile Feb 22 '12 at 18:40
  • $\begingroup$ Well, $G/H$ is a group of order $m.$ Use this to prove that $H$ is the only subgroup of $G$ which has order $p.$ $\endgroup$ – Geoff Robinson Feb 22 '12 at 18:52

Define $K = \varphi(H)$. Since $H$ is normal and $K$ is a subgroup, $HK$ is also a subgroup with order

\begin{align*}|HK| = \frac{|H||K|}{|H \cap K|}\end{align*}

Both subgroups have prime order $p$, so their intersection $H \cap K$ is trivial or of order $p$. If the intersection is trivial, then $HK$ is a subgroup of order $p^2$. This is not possible, since $p^2$ does not divide $|G|$. Thus $H \cap K$ has order $p$ and therefore $H = H \cap K =K$.


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