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changed an \ltimes to a \rtimes and added the word "this"
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Glasby
Glasby
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I take it Pablo your question can be rephrased as follows. Does there exist an epimorphism $\tau\colon A\ltimes C\to A$ where $A$ acts irreducibly on $C$ and where $\ker(\tau)\ne C$? If this is your question and $A$ can act non-faithfully, then the answer is Yes. Take As $\tau\colon{\rm C}_6\times {\rm C}_3\to {\rm C}_6$ where $\ker(\tau)={\rm C}_3\leq{\rm C}_6$. I suspect you want $A$ to act faithfully, see below.

As $K:=\ker(\tau)$ and $C$ are $A$-invariant, so is $K\cap C$. Assume $K\ne C$. Then $K\cap C=1$ by $A$-irreducibility. It follows from $C\trianglelefteq A\ltimes C$ that $\tau(C)\trianglelefteq\tau(A\ltimes C)$ or $C\trianglelefteq A$. However, $C$ is characteristically simple, by $A$-irreducibility, so it equals $T^n$ where $T$ is a finite simple group. The split extension is associated with a (nontrivial) homomorphism $\phi\colon A\to{\rm Aut}(C)$, and ${\rm Aut}(C)={\rm Aut}(T^n)$ equals ${\rm GL}_n(p)$ or ${\rm Aut}(T)\wr{\rm Sym}_n$ depending on whether or not $T$ is abelian. If $C={\rm C}_2^4$ and $A={\rm C_5}\ltimes C$ with ${\rm C}_5$ acting irreducibly, then the epimorphism $\tau\colon A\ltimes C\to A$ with $\ker(\tau)=C\leq A$ is such a "nonsplit" example. If $\phi$ is faithful, then $C=T^n\trianglelefteq A\leq T^n\ltimes{\rm Sym}_n$$C=T^n\trianglelefteq A\leq T^n\rtimes{\rm Sym}_n$ or ${\rm GL}_n(p)$. Is this the problem you wanted to explore?

I take it Pablo your question can be rephrased as follows. Does there exist an epimorphism $\tau\colon A\ltimes C\to A$ where $A$ acts irreducibly on $C$ and where $\ker(\tau)\ne C$? If this is your question and $A$ can act non-faithfully, then the answer is Yes. Take As $\tau\colon{\rm C}_6\times {\rm C}_3\to {\rm C}_6$ where $\ker(\tau)={\rm C}_3\leq{\rm C}_6$. I suspect you want $A$ to act faithfully, see below.

As $K:=\ker(\tau)$ and $C$ are $A$-invariant, so is $K\cap C$. Assume $K\ne C$. Then $K\cap C=1$ by $A$-irreducibility. It follows from $C\trianglelefteq A\ltimes C$ that $\tau(C)\trianglelefteq\tau(A\ltimes C)$ or $C\trianglelefteq A$. However, $C$ is characteristically simple, by $A$-irreducibility, so it equals $T^n$ where $T$ is a finite simple group. The split extension is associated with a (nontrivial) homomorphism $\phi\colon A\to{\rm Aut}(C)$, and ${\rm Aut}(C)={\rm Aut}(T^n)$ equals ${\rm GL}_n(p)$ or ${\rm Aut}(T)\wr{\rm Sym}_n$ depending on whether or not $T$ is abelian. If $C={\rm C}_2^4$ and $A={\rm C_5}\ltimes C$ with ${\rm C}_5$ acting irreducibly, then the epimorphism $\tau\colon A\ltimes C\to A$ with $\ker(\tau)=C\leq A$ is such a "nonsplit" example. If $\phi$ is faithful, then $C=T^n\trianglelefteq A\leq T^n\ltimes{\rm Sym}_n$ or ${\rm GL}_n(p)$. Is the problem you wanted to explore?

I take it Pablo your question can be rephrased as follows. Does there exist an epimorphism $\tau\colon A\ltimes C\to A$ where $A$ acts irreducibly on $C$ and where $\ker(\tau)\ne C$? If this is your question and $A$ can act non-faithfully, then the answer is Yes. Take As $\tau\colon{\rm C}_6\times {\rm C}_3\to {\rm C}_6$ where $\ker(\tau)={\rm C}_3\leq{\rm C}_6$. I suspect you want $A$ to act faithfully, see below.

As $K:=\ker(\tau)$ and $C$ are $A$-invariant, so is $K\cap C$. Assume $K\ne C$. Then $K\cap C=1$ by $A$-irreducibility. It follows from $C\trianglelefteq A\ltimes C$ that $\tau(C)\trianglelefteq\tau(A\ltimes C)$ or $C\trianglelefteq A$. However, $C$ is characteristically simple, by $A$-irreducibility, so it equals $T^n$ where $T$ is a finite simple group. The split extension is associated with a (nontrivial) homomorphism $\phi\colon A\to{\rm Aut}(C)$, and ${\rm Aut}(C)={\rm Aut}(T^n)$ equals ${\rm GL}_n(p)$ or ${\rm Aut}(T)\wr{\rm Sym}_n$ depending on whether or not $T$ is abelian. If $C={\rm C}_2^4$ and $A={\rm C_5}\ltimes C$ with ${\rm C}_5$ acting irreducibly, then the epimorphism $\tau\colon A\ltimes C\to A$ with $\ker(\tau)=C\leq A$ is such a "nonsplit" example. If $\phi$ is faithful, then $C=T^n\trianglelefteq A\leq T^n\rtimes{\rm Sym}_n$ or ${\rm GL}_n(p)$. Is this the problem you wanted to explore?

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Glasby
Glasby
  • 2k
  • 12
  • 22

I take it Pablo your question can be rephrased as follows. Does there exist an epimorphism $\tau\colon A\ltimes C\to A$ where $A$ acts irreducibly on $C$ and where $\ker(\tau)\ne C$? If this is your question and $A$ can act non-faithfully, then the answer is Yes. Take As $\tau\colon{\rm C}_6\times {\rm C}_3\to {\rm C}_6$ where $\ker(\tau)={\rm C}_3\leq{\rm C}_6$. I suspect you want $A$ to act faithfully, see below.

As $K:=\ker(\tau)$ and $C$ are $A$-invariant, so is $K\cap C$. Assume $K\ne C$. Then $K\cap C=1$ by $A$-irreducibility. It follows from $C\trianglelefteq A\ltimes C$ that $\tau(C)\trianglelefteq\tau(A\ltimes C)$ or $C\trianglelefteq A$. However, $C$ is characteristically simple, by $A$-irreducibility, so it equals $T^n$ where $T$ is a finite simple group. The split extension is associated with a (nontrivial) homomorphism $\phi\colon A\to{\rm Aut}(C)$, and ${\rm Aut}(C)={\rm Aut}(T^n)$ equals ${\rm GL}_n(p)$ or ${\rm Aut}(T)\wr{\rm Sym}_n$ depending on whether or not $T$ is abelian. If $C={\rm C}_2^4$ and $A={\rm C_5}\ltimes C$ with ${\rm C}_5$ acting irreducibly, then the epimorphism $\tau\colon A\ltimes C\to A$ with $\ker(\tau)=C\leq A$ is such a "nonsplit" example. If $\phi$ is faithful, then $C=T^n\trianglelefteq A\leq T^n\ltimes{\rm Sym}_n$ or ${\rm GL}_n(p)$. Is the problem you wanted to explore?