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Inner Automorphisms as a Normal Subgroup |
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| Feb26-08, 02:59 PM | #1 |
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Inner Automorphisms as a Normal Subgroup
1. The problem statement, all variables and given/known data
Let G be a group. We showed in class that the permutations of G which send products to products form a subgroup Aut(G) inside all the permutations. Furthermore, the mappings of the form [itex]\sigma_b(g)=bgb^{-1}[/itex] form a subgroup inside Aut(G) called the inner automorphisms and denoted Inn(g). Prove that the inner automorphisms form a normal subgroup of Aut(G). 2. Relevant equations None 3. The attempt at a solution I attempted this problem one way, and my professor said I was going about it the wrong way - I did multiplications of the permutations and apparently I'm supposed to use function composition of the permutations. So this is what I have: I must show that, given a permutation [itex]\tau[/itex] that [itex]\tau \sigma_b \tau^{-1}[/itex] is in the set of inner automorphisms (at least, that's how we've been showing normality in class and my professor told me that this is at least correct). So, I have [tex]\tau\left(\sigma_b\left(\tau^{-1}\left(g\right)\right)\right)&=&\tau\left(b\tau^{-1}(g)b^{-1}\right)\\ &=&\tau(b)\tau\left(\tau^{-1}(g)\right)\tau(b^{-1})\\ &=&\tau(b)g\tau(b)^{-1}\\ &=&\sigma_{\tau(b)}(g)[/tex] My question is, is [itex]\tau[/itex] equal to [itex]\tau^{-1}(g)[/itex] or is it [itex]\left(\tau(g)\right)^{-1}[/itex], or are these the same thing? I've done the problem assuming that [itex]\tau^{-1}[/itex] means [itex]\tau^{-1}(g)[/itex] and it seems to work, but I'm a bit uneasy about this. Can anybody confirm whether I've done this properly or not? |
| Feb26-08, 04:42 PM | #2 |
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| Feb26-08, 07:27 PM | #3 |
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![]() What I meant to ask was whether [itex]\tau^{-1}(g)=\left(\tau(g)\right)^{-1}[/itex] in general. I have a feeling that it's not and that's what I was uneasy about. |
| Feb26-08, 08:17 PM | #4 |
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Inner Automorphisms as a Normal Subgroup |
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