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A group $G$ is called an extension of a group $Q$ by $N$ if we have the following short exact sequence: $1 \rightarrow N \rightarrow G \rightarrow Q \rightarrow 1$. If there is a homomorphism $f: G \rightarrow N$ that is the identity on $N$, then such extensions give quite a simpler structure of $G$, namely, $G \cong N \times Q$. I was thinking of this phenomenon in the following fashion: for the group $N$, we have a group $G$ such that $N$ is a direct summand of $G$. I wondered whether such a thing happens for every nontrivial group $H$. Namely, I have the following question.

For every nontrivial group $H$, does there exist a group $G_{H}$ (depending on $H$) such that $H \unlhd G_{H}$ and if $G_{H} \cong K \times T$ for some groups $K$ and $T$, then $K \ncong H$ and $T \ncong H$?

Kan't
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Pratina
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    Sure, take $G_H$ to be the trivial group. Or a simple group not isomorphic to $H$. Are you sure you are asking the right question? – Moishe Kohan Mar 15 '24 at 16:58
  • For context, see the question from yesterday. The groups need not be additive, so $N\times Q$ is better than $N\oplus Q$. – Dietrich Burde Mar 15 '24 at 16:59
  • @Moishe Kohan Thank you for pointing it out, but after your comment, I edited my question, In the previous version, there was a small typo. – Pratina Mar 15 '24 at 17:04
  • The smallest counterexample is $S_3$. (And as Dietrich Burde said, you should write $K \times T$, not $K \oplus T$.) – Derek Holt Mar 15 '24 at 17:09
  • @Dietrich Burde Thank you for your suggestion, I have changed it. – Pratina Mar 15 '24 at 17:16
  • @Derek Holt, Can you please explain to me a little more, I know I am missing something, I did not understand why $S_{3}$ is the counterexample. – Pratina Mar 15 '24 at 17:17
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    Ih $H \unlhd G$ with $H \cong S_3$ then $G = H \times C_G(H)$. The same is true for any complete group $H$ i.e. any centreless group with trivial outer automorphism group. – Derek Holt Mar 15 '24 at 17:22

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If $H$ is an abelian group, then such a $G$ always exists.

If $H$ as a nontrivial automorphism $\varphi$, let $G=H\rtimes \langle \varphi\rangle$. This group is nonabelian, since there exists $h\in H$ such that $\varphi(h)\neq h$. Then we have $$(e,\varphi)(h,1) = (\varphi(h),\varphi)\neq (h,\varphi) = (h,1)(e,\varphi).$$ However, any group of the form $H\times\langle \varphi\rangle$ is abelian.

If $H$ does not have a nontrivial automorphism, then $H$ is cyclic of order $2$ (we need the Axiom of Choice for this if $H$ is infinite; for finite abelian groups this is easy). Then we can take $G$ cyclic of order $4$, which cannot be decomposed into a direct product with two nontrivial subgroups.

But as Derek Holt points out in comments, if $H$ is complete and nontrivial (so $Z(H)=\{1\}$ and $\mathrm{Out}(H)=\{1\}$), then no such $G$ exists. Assume that $H$ is a normal subgroup of $G$.

First we prove that $G=HC_G(H)$. Note that since $H\triangleleft G$, then $HC_G(H)$ is a subgroup.

Given any $g\in G$, $g\notin H$, we know that $g$ acts on $H$ (since $H$ is normal), so there exists $\sigma\in\mathrm{Aut}(H)$ such that $\sigma(h) = g^{-1}hg$ for all $h\in H$. Since $H$ is is complete, $\sigma$ is an inner automorphism, s0 there exists a unique $h'\in H$ such that $\sigma(h)=(h')^{-1}hh' = g^{-1}hg$. Hence for all $h\in H$, $(gh'^{-1})h(gh')^{-1}=h$.

Therefore, $g(h')^{-1}\in C_G(H)$, hence $gC_G(H) = h'C_G(H)$. Thus, $g\in HC_G(H)$. This proves that $G=HC_G(H)$.

Now we show that $H\cap C_G(H)=\{1\}$. Indeed, $H\cap C_G(H)\leq Z(H) = \{1\}$.

Finally, note that every element of $H$ centralizes $C_G(H)$, and every element of $C_G(H)$ normalizes $C_G(H)$, so $G=HC_G(H)\leq N_G(C_G(H))$. Thus, $N_G(H)\triangleleft G$.

So we have $N_G(H)\triangleleft G$, $H\triangleleft G$, $G=HN_G(H)$, and $H\cap N_G(H)=\{1\}$. Therefore, $G=H\times C_G(H)$.

Arturo Magidin
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