Talk:Aufgaben:Problem 1

The proof that \(\rho(e)=e\) is trivial, but I added it nevertheless. You will most likely lose points of you don't show it at the exam. It is very easy, though.

Lilit (talk) 12:37, 30 June 2015 (CEST)

I dont think it is necessary tho show that \(\rho\) preserves inverses for \(\rho\) to be a homomorphism, but it is certainly not wrong if you do it ;-)

Simfeld (talk) 11:57, 15 June 2015 (CEST)

I assume you want to show that \(\rho(g) \in GL(V)\) but I believe this is always implied by: G a group, V a Vector space + homomorphism. So I also don't think it is necessary.

Carl (talk) 19:15, 15 June 2015 (CEST)

Quote Wikipedia:

In mathematics, given two groups, (G, ∗) and (H, ·), a group homomorphism from (G, ∗) to (H, ·) is a function h : G → H such that for all u and v in G it holds that

\( h(u*v) = h(u) \cdot h(v) \)

where the group operation on the left hand side of the equation is that of G and on the right hand side that of H.

From this property, one can deduce that h maps the identity element e_G of G to the identity element e_H of H, and it also maps inverses to inverses in the sense that

\( h\left(u^{-1}\right) = h(u)^{-1}. \,\)

Doesn't that mean that we're already done after showing the homomorphism property?

Also, writing \(\rho(e) = e\) seems wrong to me, as it's not the same entity. One \(e\) is in \(G\), the other should be in \(GL(V)\).

--Nik (talk) 11:39, 26 July 2015 (CEST)

But we don't have a group homomorphism until we have shown that the map is from \(G\) to "\(L(V)\)" is actually into \(GL(V)\) thus into a group. Am I wrong? It is pretty trial, as it always follows for any multiplicative map from a group to the linear maps on a Vector space.

Carl (talk) 13:37, 26 July 2015 (CEST)

I guess you're right. Thanks for clarifying!

--Nik (talk) 15:45, 26 July 2015 (CEST)