Reading through Abstract Algebra, Dummit and Foote e3, it states that if $G$ is a group of order 30 and $P\in Syl_5(G)$ and $Q\in Syl_3(G)$ with one of them being normal, then both are characteristic subgroups of $PQ$; this then implies that since $PQ$ is normal, both $P$ and $Q$ have to be normal.

So then I wanted to see if it was possible to prove it in another way, more generally:

Let $PQ\unlhd G$. $WLOG$, assume $P\unlhd G$.

$\Rightarrow \forall g\in G. \forall pq\in PQ,\ g(pq)g^{-1}\in PQ$.

$\Rightarrow gp(g^{-1}g^)qg^{-1} = (gpg^{-1})(gqg^{-1})\in PQ$.

But $P\unlhd G$, thus $gpg^{-1}\in P \Rightarrow gqg^{-1}\in Q$.

$\therefore Q\unlhd G$.

Does this work? Thank you!

  • $\begingroup$ The title of your post is not wonderful, because the statement there is not true in general. $\endgroup$
    – Derek Holt
    Commented May 13, 2017 at 19:22

1 Answer 1


The claim is false. For example, in $\;S_4\;$ , take

$$P:=\left\{\,(1),\,(12)(34),\,(13)(24),\,(14)(23)\,\right\}\lhd S_4\;,\;\;Q=\langle\,(123)\,\rangle=\{\,(1),\,(123),\,(132)\,\}\le S_4$$

Then $\;PQ=A_4\lhd S_4\;$ , yet $\;Q\;$ is not normal in $\;S_4\;$ , nor in $\;A_4\;$ .

I'm not sure to what part of D&F you're referring (a chapter, section note will help), but in the case of a group of order $\;30\;$ a subgroup of order $\;15\;$ is always normal as its index is $\;2\;$ ...and there are groups of order $\;30\;$ whose groups of order $ \;3\;$ are not normal, for example.

  • 1
    $\begingroup$ I found it in Chapter 4, Section 5 under groups of order 30. Thanks for the concrete example! $\endgroup$
    – kdavid2
    Commented May 13, 2017 at 17:35
  • $\begingroup$ @kdavid2 Any time. Groups of order $\;30\;$ are a special in this respect. $\endgroup$
    – DonAntonio
    Commented May 13, 2017 at 17:39

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