Timeline for Why $G/F(G)$ is isomorphic to a subgroup of ${\rm Out}(F(G))$?
Current License: CC BY-SA 4.0
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| May 7, 2020 at 15:38 | vote | accept | CommunityBot | ||
| May 6, 2020 at 12:36 | vote | accept | CommunityBot | ||
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| May 6, 2020 at 11:46 | comment | added | user121195 | @GeoffRobinson Yes, I was confused by use of the third isomorphism there. What we need to apply here is actually the first isomorphism theorem. Thanks! I’ll write an answer and make it clear that the third isomorphism theorem here is not really necessary. | |
| May 6, 2020 at 11:42 | comment | added | Geoff Robinson | That's what I explained (or tried to) in the comments before. These are now deleted. It's indeed true that the third isomorphism theorem isn't really necessary. | |
| May 6, 2020 at 11:34 | comment | added | user121195 | @GeoffRobinson When I went through the proof, I found we actually don’t need to use the third isomorphism theorem: Because $F(G)$ is normal in $G$, there’s a homomorphism $\phi:G\to {\rm Aut}(F(G))$ whose kernel is $C_G(F(G))=Z(F(G))\subseteq F(G)$. We have $G/F(G)\mathbf{\cong \phi(G)/\phi(F(G))=}\phi(G)/{\rm Inn}(F(G))\le {\rm Aut}(F(G))/{\rm Inn}(F(G))$. So although the isomorphism theorem actually fails to help, the real key part is $G/F(G)\cong \phi(G)/\phi(F(G))$ which is actually linked to the first isomorphism theorem. Please tell me if my argument does not make sense. Thanks. | |
| May 5, 2020 at 3:25 | comment | added | Arturo Magidin | @Hello: I understand you are not able to delete the question. Nonetheless, it was important to indicate the cross post with links to prevent duplication of efforts. | |
| May 5, 2020 at 3:24 | comment | added | user121195 | @ArturoMagidin Sorry for that. I won’t do that again. And thanks for your contribution to both sites! | |
| May 5, 2020 at 3:22 | comment | added | user121195 | @ArturoMagidin thanks. I was about to delete this post but there was an answer then so I wasn’t able to delete it. Now I know that one should not post the same questions both on MO and MSE, sorry for that. I just thought MO and MSE are not the same site, and many contributors don’t visit both MO and MSE, so I asked a couple of questions that I thought suitable for both sites. I’ve deleted all that kind of posts if I’m able to. You mentioned that I’ve been a user for more that two years. Yes, but actually I had not been using this site for about one and half years before the beginning of 2020. | |
| May 5, 2020 at 3:12 | comment | added | Arturo Magidin | Posted nigh-simulatenously in math.stackexchange: math.stackexchange.com/questions/3655320/… | |
| May 3, 2020 at 4:27 | answer | added | Santana Afton | timeline score: 5 | |
| May 2, 2020 at 20:05 | review | Close votes | |||
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| May 2, 2020 at 16:25 | comment | added | Geoff Robinson | It is not a question of "fault", but you do probably have to figure it out for yourself. | |
| May 2, 2020 at 16:15 | comment | added | user121195 | @GeoffRobinson Thanks for your patience. You are very kind. It’s my fault and I will think more about it. | |
| May 2, 2020 at 16:12 | history | edited | user121195 | CC BY-SA 4.0 |
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| May 2, 2020 at 16:04 | history | edited | user121195 | CC BY-SA 4.0 |
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| May 2, 2020 at 15:56 | comment | added | user121195 | @GeoffRobinson I want to take an example. Please point it out if it doesn’t make sense. Assume that $A$ is a subgroup of $G$ and $B\trianglelefteq A$. Also, assume that $N\trianglelefteq M$. If $A\cong M$ and $B\cong N$, then it is Not true in general that $A/B\cong M/N$. So in the case that we were talking about, $G={\rm Aut}(F(G))$, $B={\rm Inn}(F(G))$, $M=G/Z(F(G))$, $N=F/Z(F(G))$, it’s just the same: $M$ is isomorphic to a subgroup of $G$, namely $A$, and $N\cong B$. But we don’t have $A/B\cong M/N$ in general. What I was asking was how to prove it in this specific case. | |
| May 2, 2020 at 15:39 | history | edited | user121195 | CC BY-SA 4.0 |
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| May 2, 2020 at 15:35 | comment | added | user121195 | @GeoffRobinson Thanks for your kind comments. Actually those pieces were all mentioned in my original post and I’m afraid that they are not what I was asking. It really seems close to the conclusion with those pieces. But I just don’t know how to push forward. It’s quite obvious that $G/F(G)\cong G/Z(F(G)\big/ F(G)/Z(F(G))$ is isomorphic to a subgroup of ${\rm Aut}(G)/{\rm Inn}(G)$ since the $G/Z(F(G))$ is isomorphic to a subgroup of ${\rm Aut}(G)$ and $F(G)/Z(F(G))\cong {\rm Inn}(G)$. But it’s not sufficient. I think there’s still something missing, I also mentioned that in my post. | |
| May 2, 2020 at 15:20 | history | edited | user121195 | CC BY-SA 4.0 |
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| May 2, 2020 at 15:12 | comment | added | user121195 | @GeoffRobinson Thanks. You mean $G/F(Z(G))$ is isomorphic to a subgroup of $ {\rm Aut}(F(G))$. Am I right? I know this. It can be proved by $N/C$ theorem. I write that in my original post. It’s not exactly where I got stuck. I noticed there was a typo and it might have made you think I think that map is surjective. I know $G/F(Z(G))$ is isomorphic to a subgroup of ${\rm Aut}(F(G))$ and $F(G)/Z(F(G))\cong {\rm Inn}(F(G))$ and I also know I can apply the third isomorphism theorem to get closer to the conclusion. But I just don’t know what to do next. I wonder if my question is not clear. | |
| May 2, 2020 at 15:06 | history | edited | user121195 | CC BY-SA 4.0 |
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| May 2, 2020 at 14:27 | comment | added | user121195 | @GeoffRobinson $F(G)/Z(F(G))\cong {\rm Inn}(F(G))$ and by (1) that $G/Z(F(G))\cong {\rm Aut}(F(G))$. If it is true that, say, “if $A\cong M$ and $B\cong N$ where $B\trianglelefteq A$ and $N\trianglelefteq M$ then $A/B\cong M/N$”, then we’re done. However, it is not true in general. It’s exactly where I got stuck. The isomorphism theorem does bring me very close to the conclusion. But there’s still something in my way. | |
| May 2, 2020 at 14:22 | comment | added | user121195 | @GeoffRobinson Thanks. I mentioned that in my post, this isomorphism theorem does play a key role in the proof. But I got stuck after using this theorem. | |
| May 2, 2020 at 14:18 | comment | added | YCor | Often "X is isomorphic a subgroup of Y" is poor way of starting "this natural map from X to Y is an isomorphism", which in case is not easy to state can better be stated as "there is a natural injective homomorphism from X to Y". By poor way, I mean that when one applies such a statement, one needs to refer to some given natural map, not the bare existence of such a map. | |
| May 2, 2020 at 14:11 | history | asked | user121195 | CC BY-SA 4.0 |