Timeline for Automorphism group of a free product
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2015 at 10:22 | vote | accept | Ofra | ||
| Nov 18, 2015 at 10:20 | answer | added | HJRW | timeline score: 11 | |
| Nov 18, 2015 at 10:19 | history | edited | Ofra | CC BY-SA 3.0 |
deleted 5 characters in body; edited title
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| Nov 18, 2015 at 10:12 | comment | added | HJRW | @QiaochuYuan, your example isn't a great one, since $\mathrm{Out}(\mathbb{Z}/2)=1$ and $\mathrm{Out}(\mathbb{Z}/3)=\mathrm{Out}(PSL_2(\mathbb{Z}))=\mathbb{Z}/2$, so in the outer automorphism group does in fact arise in a simple way from the outer automorphism groups of the factors. | |
| Nov 18, 2015 at 8:10 | comment | added | YCor | It's called "free product". It is also the amalgam (or amalgamated product, but not amalgamate) of the two groups over the trivial subgroup. | |
| Nov 18, 2015 at 8:08 | answer | added | Thomas | timeline score: 6 | |
| Nov 18, 2015 at 6:51 | comment | added | Sh.M1972 | Consider the case $G=H=\mathbb{Z}$. Then we have $Aut(G)=Aut(H)=\mathbb{Z}_2$, but $G\ast H=F_2$, which has a very complicated automorphism group. | |
| Nov 18, 2015 at 4:25 | comment | added | Qiaochu Yuan | It is certainly not a straightforward function of $\text{Aut}(G)$ and $\text{Aut}(H)$. For example, take $G = \mathbb{Z}_2, H = \mathbb{Z}_3$. Then $G \ast H \cong PSL_2(\mathbb{Z})$, whose automorphism group is, I imagine, $PGL_2(\mathbb{Z})$. | |
| Nov 18, 2015 at 2:37 | comment | added | Alex Kruckman | Surely this depends on the particular $G$ and $H$... | |
| Nov 18, 2015 at 1:46 | history | asked | Ofra | CC BY-SA 3.0 |