Difference between revisions of "Talk:Aufgaben:Problem 3"
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 21:24, 20 July 2015 (CEST) | [[User:Carl|Carl]] ([[User talk:Carl|talk]]) 21:24, 20 July 2015 (CEST) | ||
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| + | In ''d)2.'': Why do we need to show that "\(\{L_g: g \in G\}\) is a group under composition"? | ||
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| + | After ''b)'', we already know that \(L\) is a map with domain \(G\) and codomain \(\mathrm{Sym}G\) and from''c)'' we know that both of these are groups. Shouldn't it be enough to show the homomorphism property? | ||
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| + | --[[User:Nik|Nik]] ([[User talk:Nik|talk]]) 10:04, 29 July 2015 (CEST) | ||
Revision as of 08:04, 29 July 2015
Q: In Part c) 1. 2. associativity is used to prove it is well defined. And for proving associativity we need it to be well defined. This seems like a problem, how dowe solve it?
A: ?
Is not 1.1 the proof of the group operation being well defined. And 1.2 the proof of closure? I can't see why we could't first show 1.1 then associativity and then the rest. Why would we need closure to prove associativity? Associativity for maps follow directly from the definition of composition, if the domains and images fit together, which is clear after 1.1.
Carl (talk) 21:24, 20 July 2015 (CEST)
In d)2.: Why do we need to show that "\(\{L_g: g \in G\}\) is a group under composition"?
After b), we already know that \(L\) is a map with domain \(G\) and codomain \(\mathrm{Sym}G\) and fromc) we know that both of these are groups. Shouldn't it be enough to show the homomorphism property?
