If $H$ and $K$ are supersolvable, normal subgroups of $G$, does it follow that $HK$ is supersolvable?

Question

Let $H$ and $K$ be supersolvable, normal subgroups of some group $G$. Does it follow that $HK$ is supersolvable?

This is true if $H \cap K = 1$ since a direct product of two supersolvable groups is supersolvable. However, I think counterexamples for the statement exist when $H \cap K \neq 1$. What is the smallest (finite) example? Any example that is particularly easy to prove?

score 5 · Accepted Answer · answered Aug 04 '13 at 22:53

5

The smallest example is $G=S_3 \wr S_2$, the wreath product of $S_2$ acting on two copies of $S_3$. Take $M$ to be the base group, $$M=\langle (1,2), (5,6), (1,2,3), (4,5,6) \rangle = S_3 \times S_3,$$ and $N$ to be a twisted copy: $$N=\langle (1,2)(4,5), (1,4)(2,5)(3,6), (1,2,3), (4,5,6) \rangle \cong S_3 \times S_3.$$

Then $MN=G$ has a chief factor $A_3 \times A_3$ which is not simple, so $G$ is not supersolvable.

answered Aug 04 '13 at 22:53

Jack Schmidt

56,967

How does the chief factor $A_3\times A_3$ not being simple imply $G$ is not supersolvable? – Shaun Jan 22 '22 at 12:28
I have asked this question here. – Shaun Jan 22 '22 at 18:50

If $H$ and $K$ are supersolvable, normal subgroups of $G$, does it follow that $HK$ is supersolvable?

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