Talk:Aufgaben:Problem 3
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?
--Nik (talk) 10:04, 29 July 2015 (CEST)
I believe that part should be in e) and not in d). Because \(L :G \rightarrow \mathrm{Sym}G\) is not surjective we take the restriction \(L :G \rightarrow \mathrm{Im}G = \{L_g: g \in G\}\) and have to show that it is a subgroup of \(\mathrm{Sym}G\).
Djanine (talk) 11:16, 29 July 2015 (CEST)
I agree with Djanine. This is how I would solve it:
d) L is homomorphism (see solution). L is group homomorphism G -> SymG. (Since G is a group by definition and in c) we showed that SymG has group structure)
e) to show: Every finite group can be considered a subgroup of SymG. In other words: Every finite group G is isomorphic to \( \mathrm{Im}G = \{L_g: g \in G\}\), which is a subgroup of SymG.
1. \(\{L_g: g \in G\}\) is a subgroup of SymG. (see solution)
2. In b) we showed that L: G -> SymG is injective, but its not surjective. If we restrict L to its image, we get a map \(L^* :G \rightarrow \mathrm{Im}G = \{L_g: g \in G\}\). This new map is still injective, but also surjective by definition, and therefore bijective. \(L^*\) is then a group isomorphism G -> ImG, and every finite group G is isomorphic to ImG.
-- Snick (talk) 13:46, 4 August 2015 (CEST)
ok I'll fix that
