# Normal products and radicals in finite groups

If $G$ is a finite group with normal subgroups $M$ and $N$, then $MN$ is a subgroup, called the normal product of $M$ and $N$.

If $\mathcal{F}$ is a set of finite groups closed under isomorphism and normal products, then there is a subgroup $O_\mathcal{F}(G)$ called the $\mathcal{F}$-radical which is the normal product of all normal subgroups of $G$ that are in $\mathcal{F}$. If $\mathcal{F}$ additionally is closed under normal subgroups, $O_\mathcal{F}(G)$ has the property that a subgroup of $G$ is a subnormal subgroup from $\mathcal{F}$ if and only if it is a subnormal subgroup of $O_\mathcal{F}(G)$. This is standard material on Fitting classes, covered in say 6.3 of Kurzweil–Stellmacher's textbook.

I'd like to conclude quite generally that $O_\mathcal{F}(MN) = O_\mathcal{F}(M) O_\mathcal{F}(N)$. This is definitely true of direct products, and I'm having a hard time seeing any relevant differences to the normal product. However, I am also having trouble proving it.

I get that $O_\mathcal{F}(M) O_\mathcal{F}(N) \leq O_\mathcal{F}(MN)$ in general, simply because $\mathcal{F}$ is closed under normal products.

How do I show the reverse containment? If it is not true, is there some extra hypothesis on $\mathcal{F}$ that does it (like subgroup or quotient closure)?

The particular case of concern is $\mathcal{F}$ consisting of all $\pi$-groups. I can brute-force it for $\mathcal{F}$ consisting of all solvable $\pi$-groups, and technically all groups my audience was considering were solvable, but I'd prefer a technique that worked in general, or some counterexamples to show what extra hypotheses are actually being used.

For instance, in my application $\mathcal{F}$ is closed under quotients and all subgroups too (a subgroup closed (saturated by Bryce-Cossey) Fitting formation), but I doubt much if any of that extra hypothesis is needed.

Edit: Assume $\mathcal{F}$ is quotient closed for the positive answer. I don't current have a counterexample for the general $\mathcal{F}$, but they are apparently plentiful and well known.

Apparently direct products don't work this way for general $\mathcal{F}$ (for any normal Fitting class properly contain in the class of all solvable groups, for instance). Lockett figured out how to fix this in a fairly low impact way. For any Fitting class $\mathcal{F}$, he associated $\mathcal{F}^*$ with the property that $\mathcal{F} \subseteq \mathcal{F}^* = \mathcal{F}^{**} \subseteq \mathcal{F} \mathcal{A}$ and that $O_{\mathcal{F}^*}(G \times H) = O_{\mathcal{F}^*}(G) \times O_{\mathcal{F}^*}(H)$ and $O_{\mathcal{F}^*}(G) = \{ g \in G : (g,h) \in O_\mathcal{F}(G\times G) \}$.

Theorem 2.2d shows that if $\mathcal{F}$ is closed under quotients or residual products (the other aspect of being a formation), then $\mathcal{F}=\mathcal{F}^*$, so all Fitting formations work the way I thought.

... Still checking on normal products not directly addressed in the article ...

• Lockett, F. Peter. “The Fitting class $\mathfrak{F}^*$” Math. Z. 137 (1974), 131–136. MR364435 DOI:10.1007/BF01214854
• There is some small possibility that this is not true, and the class I'm looking for are called Lockett classes, but that is a few hundred pages of reading ahead of me, so I'm very happy for any summaries. – Jack Schmidt Jun 8 '13 at 22:53
• Normal Product? You should have answered my question here! – user1729 Aug 1 '13 at 19:01

Let $\mathcal{F}$ be the class of all $2$-groups. Consider $G = S_3 \times C_2$. For our normal subgroups $M$ and $N$, let $M = S_3 \times 1$ and $N = \{(\sigma, \operatorname{sgn}(\sigma)): \sigma \in S_3\}$. Now $M \cong N \cong S_3$, so $O_{\mathcal{F}}(M)O_{\mathcal{F}}(N)$ is trivial. But $O_{\mathcal{F}}(MN) = O_{\mathcal{F}}(G) = 1 \times C_2$.
Imitating the above example, I think the following should also work. Let $\mathcal{F}$ be a class of finite groups closed under isomorphism and normal products, and such that $O_{\mathcal{F}}(S_n)$ is trivial and $O_{\mathcal{F}}(C_2) = C_2$ (ie. the cyclic group of order $2$ is a $\mathcal{F}$-group). Like in the previous example, let $G = S_n \times C_2$ and $M = S_n \times 1$ and $N = \{(\sigma, \operatorname{sgn}(\sigma)): \sigma \in S_n\}$. Then $M$ and $N$ are normal in $G$ and $M \cong N \cong S_n$ so $O_{\mathcal{F}}(M)O_{\mathcal{F}}(N)$ is trivial. Now $1 \times C_2$ is a nontrivial normal $\mathcal{F}$-subgroup of $G = MN$, so $O_{\mathcal{F}}(MN)$ is nontrivial.
• Thanks, $O_\mathcal{F}$ in your first example is called $O_2$ and is of fundamental importance. I am disturbed to have not seen this before. Thanks! – Jack Schmidt Aug 1 '13 at 18:03
• Same group works for $\mathcal{F}$ the class of 3-nilpotent groups. Fixed in the 3-length 1 proof. – Jack Schmidt Aug 1 '13 at 18:22