Timeline for On some endomorphisms of finite groups of odd order
Current License: CC BY-SA 3.0
9 events
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| Jan 11, 2014 at 12:13 | comment | added | zibadawa timmy | This doesn't seem right at all. $\phi(A)=A$ and $\ker(\phi)\leq A$ is impossible for $A$ finite. And there is no guarantee that the image of a higher power of $\phi$ and $K$ will generate $G$. And in your last sentence you can't fix $K$ and invert every element of it at the same time. Unless it has exponent 2, where those are the same. | |
| Oct 30, 2013 at 18:22 | vote | accept | Yassine Guerboussa | ||
| Oct 3, 2013 at 20:40 | comment | added | Yassine Guerboussa | Yes I see. I'm confused when you are saying $\phi$ fixes $A$, you just meant that $A$ is $\phi$-invariant. Note that this is included in our statement that $\phi$ acts trivially on $G/A$. | |
| Oct 3, 2013 at 20:33 | history | edited | Derek Holt | CC BY-SA 3.0 |
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| Oct 3, 2013 at 20:25 | history | edited | Derek Holt | CC BY-SA 3.0 |
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| Oct 3, 2013 at 20:08 | comment | added | Derek Holt | Yes, so $x \in A$ and hence $K \le A$, which is what I wrote. Why should that imply $x=1$? I didn't say that $\phi$ acts trivially on $A$. I am proving the contrapositive of what you asked. I am assuming that there is an endomorphism of $G$ that acts trivially on $G/A$ that is not an automorphism (which is certainly possible in general!) and deducing that there must be an automorphism of $G$ that induces the identity on $G/A$ but not on $A$. | |
| Oct 3, 2013 at 16:53 | comment | added | Yassine Guerboussa | You said $\phi$ is an endomorphism that acts trivially on $G/A$ and $A$ and has a non trivial kernel. This can not hold, as $\phi(x)=1$ implies $x^{-1} \phi(x) = x^{-1} \in A$, thus $x \in A$ and so $\phi(x)=x=1$. | |
| Oct 3, 2013 at 15:53 | history | edited | Derek Holt | CC BY-SA 3.0 |
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| Oct 3, 2013 at 15:48 | history | answered | Derek Holt | CC BY-SA 3.0 |