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This is concerning (part of) Exercise 5.4.5 and Exercise 5.4.6 of Robinson's, "A Course in the Theory of Groups (Second Edition)".

I have done the first one; the second might take me a while.

The Details:

The dihedral group is given by the presentation

$$D_{2n}\cong \langle r,s\mid r^n, s^2, srs=r^{-1}\rangle.$$

On page 150, paraphrased,

A group $G$ is supersolvable if it has a series of normal subgroups

$$1=G_0\unlhd G_1\unlhd \dots\unlhd G_n=G$$

all of whose factors $G_{i+1}/G_i$ are cyclic.

On page 157, we have the two exercises: part of

Exercise 5.4.5: The class of supersolvable groups is closed with respect to [. . .] finite direct products

. . . and . . .

Exercise 5.4.6: The product of two normal supersolvable subgroups need not be supersolvable. [Hint: Let $X$ be the subgroup of $GL(2,3)$ generated by

$$\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}\text{ and }\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix},$$

thus $X\cong D_8$. Let $X$ act in the natural way on $A=\Bbb Z_3\oplus\Bbb Z_3$ and write $G=X\ltimes A$. Show that $G$ is not supersolvable. Let $L$ and $M$ be distinct Klein $4$-subgroups of $X$ and consider $H=LA$ and $K=MA$.]

The Question:

How does Exercise 5.4.5 not contradict Exercise 5.4.6?

Thoughts:

In Exercise 5.4.6, does Robinson mean the subgroup product

$$ST=\{ st\in C\mid s\in S, t\in T\}$$

for subgroups $S, T$ of a group $C$? I think so.

If not, then why does it not contradict Exercise 5.4.5?

A solution to (this part of) Exercise 5.4.5 can be found here:

Please help :)

Shaun
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    The exercise says: "The product of two normal supersolvable subgroups ...", and then you ask "does Robinson mean the subgroup product?". Isn't that exactly what it says? – Derek Holt Jan 20 '22 at 16:41
  • I suppose so, @DerekHolt; it just wasn't clear to me at first. Thank you. – Shaun Jan 20 '22 at 16:44
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    There is no ambiguity about the product of two subgroups of the same group. But occasionally people talk about the product of two groups, where there is no connection between the two groups. That is very unclear. Direct product? Free product? – Derek Holt Jan 20 '22 at 17:12
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    Why do you think they contradict each other? 5.4.6 says that if $G$ is a group, and $H,K$ are both normal in $G$ and each is supersolvable, it need not be the case that $HK$ (which is a subgroup of $G$) is supersolvable (I would also call this "the product of the subgroups", but perhaps that's just me). 5.4.4 says that if $G_1,\ldots,G_n$ are each supersolvable, then $G_1\times\cdots\times G_n$ is supersolvable. Note that for $H$ and $K$ as above, $HK$ need not be isomorphic to a direct product. – Arturo Magidin Jan 21 '22 at 15:58
  • Thank you. My issue was I thought that perhaps some internal direct product was intended, @ArturoMagidin. I suppose this was a stupid question . . . – Shaun Jan 21 '22 at 16:04
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    @Shaun: No worries. We all do that kind of thing from time to time: can't possibly be that, I must be missing something.... ooops, not, it really is that. :-) – Arturo Magidin Jan 21 '22 at 16:14
  • This answer is an example of $ST$ not being supersolvable for normal, supersolvable subgroups $S$ and $T$. – Shaun Jan 21 '22 at 20:52

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