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Why do non-abelian characteristically simple groups have no nontrivial subnormal subgroups of prime power order?

Let $N$ be a finite non-abelian characteristically simple group.

A finite group is characteristically simple if and only if it is the direct product of isomorphic simple groups. We can know that $N$ is the direct product of some isomorphic non-abelian simple groups. Therefore $N$ cannot be the direct product of some $p$-groups.

But I don’t know what to do next. Any help is appreciated.

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    $\begingroup$ The simple answer to your question is that such a subgroup would be contained in $O_p(G)$, so $G$ would be a $p$-groups, etc. $\endgroup$ – Derek Holt Jan 28 '20 at 9:00
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    $\begingroup$ Any subnormal $p$-subgroup of any finite group $G$ is contained in $O_p(G)$. That's because $O_p(G)$ is a characteristic subgroup of $G$. $\endgroup$ – Derek Holt Jan 28 '20 at 12:27
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Technically, the trivial group has prime power order and is certainly subnormal (normal, even); so you need to exclude it. You mean nontrivial subnormal subgroups of prime power order.

Claim: If $S$ is nonabelian and simple, then the only normal subgroups of $S^n = \underbrace{S\times \cdots \times S}_{n\text{ factors}}$ are the subgroups of the form $H_1\times\cdots\times H_n$, with $H_i=\{e\}$ or $H_i=S$ for each $i$.

To see this, let $N$ be normal in $S^n$. The projection onto the $i$th component is normal in the image, hence is either trivial or all of $S$. Thus, we may reduce to the case in which $N$ is a normal subgroup of $S^n$ and all projections equal $S$.

Now consider the intersection of $N$ with $S\times\{e\}\times\cdots\times\{e\}$. This is normal in the latter subgroup, which is isomorphic to $S$; thus, this intersection is either trivial or all of $S$. Let $g\in S$, $g\neq e$. Since the projection of $N$ onto the the first coordinate is all of $S$, there exist $g_2,\ldots,g_n\in S$ such that $(g,g_2,\ldots,g_n)\in N$. Since $S$ is simple and nonabelian, there exists $x\in S$ such that $xgx^{-1}\neq g$. Since $N$ is normal, $$(x^{-1},e,\ldots,e)(g,g_2,\ldots,g_n)(x,e,\ldots,e) = (x^{-1}gx,g_2,\ldots,g_n)\in N.$$ Multiplying on the left by $(g,g_2,\ldots,g_n)^{-1}$ we get that $(g^{-1}x^{-1}gx,e,\ldots,e)=([g,x],e,\ldots,e)\in N$. Thus, the intersection of $N$ with $S\times\{e\}\times\cdots\times\{e\}$ is nontrivial, and by normality must be all of $S\times\{e\}\times\cdots\times\{e\}$.

Similar arguments show that the intersection of $N$ with each coordinate subgroup is the whole coordinate subgroup, hence $N=S^n$. This proves the claim.

So now assume that $P$ is subnormal in $S^n$. Then there exist a sequence of subgroups $N_1\subseteq N_2\subseteq\cdots\subseteq N_k = S^n$ such that $P=N_1$, $N_1\triangleleft N_2,\ldots, N_{k-1}\triangleleft N_k=S^n$. But each of the $N_i$ is just a direct product of copies of $S$; in particular, $N_2$ is a direct product of copies of $S$, and $P$ is normal in $N_2$, hence is a direct product of copies of $S$. But since $S$ is simple and nonabelian, $P$ is not a $p$-group unless it is trivial.

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  • $\begingroup$ @Math: As it says at the very top, $S$ is a nonabelian simple group. A characteristicly simple group is isomorphic to the direct product of finitely many copies of the same nonabelian simple group; otherwise, taking the factors that correspond to a particular isomorphism class would give you a characteristic subgroup. $\endgroup$ – Arturo Magidin Jan 28 '20 at 15:02
  • $\begingroup$ @Math: I reduced to the case where all projections are $S$;if any projection is trivial, you know that $N$ is just trivial in that coordinate and it can be ignored. $\endgroup$ – Arturo Magidin Apr 16 '20 at 1:20
  • $\begingroup$ @Math: Two and a half months later and only now do you get around to asking what “the projection” means? You’ve got to be kidding me... Given a direct product $\prod_{i\in I}G_i$, the projections are the maps $\pi_j\colon \prod_{i\in I}G_i\to G_j$ that map the tuple $(g_i)_{i\in I}$ to the $j$th coordinate $g_j$. $\endgroup$ – Arturo Magidin Apr 16 '20 at 3:56
  • $\begingroup$ @Math: Any coordinate where the projection is trivial is just trivial always. All elements of $N$ just have $e$ in that coordinate. We know what happens in that coordinate, that coordinate is irrelevant. If you have $N\leq S\times S$, and you know that the second coordinate of every element of $N$ is $e$, then $N$ is really just a subgroup of the first $S$, with an “$e$” tacked on at the end. Same here. We know what happens in each coordinate where the projection is trivial, so we just need to worry about what happens in the other coordinates. $\endgroup$ – Arturo Magidin Apr 16 '20 at 3:57
  • $\begingroup$ @Math: And by the way: why on Earth did you accept this answer back in January if you have so little idea about what it actually means? That seems like a less than smart way to use the “accept” function. $\endgroup$ – Arturo Magidin Apr 16 '20 at 3:58

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