I am still working on this problem on radical of finite group:

Assume that $R(G)$ is simple and not commutative, show that $G$ is a subgroup of $Aut(R(G))$.

I have managed to parse the problem into these followings, consisting of what I know and what I don't know:

(1) In my class note the radical of finite group $R(G)$ is defined as: $$R(G) := E(G)F(G),$$ where $E(G)$ and $F(G)$ are called respectively the Layer and Fitting of $G$. Unfortunately, the definitions of $E(G)$ and $F(G)$ are long, winding and arduous chains of sub-definitions, which I believe won't be useful for solving this problem. Fortunately, the same text has established that $E(G)$ and $F(G)$ are normal subgroup of $G.$ Hence it is easy to conclude that $R(G) \lhd G$.

(2) The problem states that $R(G)$ is simple, meaning that $R(G)$ does not have any non-trivial normal subgroup. It is further stated that $R(G)$ is non-ablien.

(3) Now, $Aut(R(G))$ is group of isomorphisms from $R(G)$ to itself. I learn from Wikipedia here that conjugation is a handy example of automorphism. Therefore I visualize the $Aut(R(G))$ as follow as an aid in solving this problem: $$\begin{align} \varphi_g &: R(G) \to R(G) \qquad &&g \in G\\ &: r \mapsto r^g &&r \in R(G) \\ &: r \mapsto grg^{-1} \end{align}$$ Please correct me if there is any misstep in this visualization.

(4) Now, here comes the biggest challenge for me: How to prove $G$ is a subgroup of $Aut(R(G))$? Especially, how to prove, first and foremost, that $G$ is a subset of $Aut(R(G))$? To me at least, this proposition is counter-intuitive since $R(G)$ is normal subgroup of $G$. Did I miss anything here?

Any help or take from you would be most appreciated. Thank you for your time and help.

POST SCRIPT AFTER RESPONSE FROM "mesel" ~~~~~~~~~~~~~~
Thanks to "mesel" for his deep analysis, and here is the line-by-line analysis as I understood it. Any misunderstanding herein, though, will completely be mine.

(1) From the Fundamental Lemma of Finite Group Theory, we have $C_G(R(G)) \subseteq R(G)$, and from that $C_G(R(G)) \leq R(G)$ easily follows. Since $C_G(R(G))$ is normal, therefore it is the normal subgroup of $R(G)$.
(2) But the question states that $R(G)$ is simple, meaning $C_G(R(G))$ is either $R(G)$ itself or $e$.
(3) If $R(G) = C_G(R(G))$, then $R(G)$ must be abelian which violates the premise given by the problem, therefore $C_G(R(G)) = e$.
(4) Notice that $C_G(R(G)) = \{g \in G \mid gr = rg, \forall r \in R(G) \}$, implying that
$$\begin{align} gr &= rg \\ g &= rgr^{-1} \\ &= e. \\ \end{align}$$ (5) Let $\phi$ be a homomorphism from $G$ to $Aut(R(G))$, where the automorphism is a conjugation: $$\begin{align} \phi &: G \to \underbrace{(R(G) \to R(G))}_{Aut(R(G)} \\ &: \underbrace {g}_{= \ e} \mapsto (r \mapsto \underbrace {rgr^{-1}}_{= \ e}) \qquad \qquad \forall r \in R(G), \\ \end{align}$$ which implies that $\phi$ is monomorphism.
(6) Because of the injective homomorphism above, $G \cong \phi (G)$, and since $\phi(G) \leq Aut(R(G))$, therefore we conclude that $G$ is subgroup of $Aut(R(G))$ as required. $\blacksquare$


Lemma1: $C_G(R(G))\leq R(G)$. (property of $R(G)$)

Lemma2: If $N$ is a normal subgroup of $G$ then $C_G(N)$ is also a normal subgroup of $G$.

By lemma1,2 we can say that $C_G(R(G))$ is a normal subgroup of $G$ which is contained in $R(G)$.

But since $R(G)$ is simple, $C_G(R(G))=1$ or $C_G(R(G))=R(G)$.

ıf $C_G(R(G))=R(G)$ then $R(G)$ is abelian which is not the case.

Thus we have $C_G(R(G))=e$.

Now, Let $G$ act on $R(G)$ by conjugation then you will get homomorphism $\phi$ from $G$ to $Aut(R(G))$ with kernel $C_G(R(G))=e$. Then $\phi$ is an embedding as kernel is trivial. We are done.

  • $\begingroup$ Thanks for your response! I can pretty much digest your analysis albeit slowly, but I have difficulties understanding the last paragraph. Do you mind if you elaborate it into line-by-line? Especially: (1) How do you know that $C_G(R(G))$ is the kernel of $G$? (2) What is an "embedding"? And finally, (3) How do you go from proposition that $\phi$ is an embedding to conclude that $G$ is subgroup of $Aut(R(G))$? $\endgroup$ – Amanda.M Jan 7 '15 at 20:45
  • $\begingroup$ @A.Magnus: If a group $G$ act on a normal subgroup $N$ by conjugation then $ker=\{g\in G| gng^{-1}=n \ for \ all \ n\}$. Notic that $Ker$ is the exactly $C_G(N)$. If you take $R(G)=N$ then the result fallows. When the kernel is trivial, $\phi$ is one to one homomorphism. Thus, $\phi(G)\leq Aut(G)$ and $G\cong \phi(G)$ which conclude the result. $\endgroup$ – mesel Jan 7 '15 at 21:30
  • $\begingroup$ Many thanks to you! I parsed out your analysis line-by-line in the Post Script above. But anyway, don't you think that $ker(\phi)$ should be $\{g \in G \mid gng^{-1} = e, \forall n \in N \}$ instead? Also, shouldn't it be $\phi(G) \leq Aut(R(G))$ instead? Thanks again. $\endgroup$ – Amanda.M Jan 8 '15 at 16:59
  • $\begingroup$ @A.Magnus: The kernel is consisting of elements beheving likes identity function. Thus, $Ker(\phi)=\{g\in G|gng^{-1}=n \}$. (I think you confused the term behaving like identity function and sending elements to identity). By the way, You are welcome. $\endgroup$ – mesel Jan 8 '15 at 17:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.