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 July 7th, 2014, 12:11 PM #1 Newbie   Joined: Jul 2014 From: Europe Posts: 1 Thanks: 0 Direct sum of Hom-Sets I am struggling to prove the following statement: Hom(V,W_1 + W_2) = Hom(V,W_1) + Hom(V,W_2), where + denotes the direct sum, V, W_1 and W_2 are G-modules and V is irreducible. Hom(X,Y) denotes the set of homomorphisms from X to Y. I have already looked elsewhere, but I just found explanations involving category theory that I do not understand. Is there an easy explanation, why the statement is true? Thank you very much for your help! July 7th, 2014, 02:29 PM #2 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 I've never seen this before, but I bet I can figure it out. Suppose we have the maps: $\pi_1:W_1\oplus W_2 \to W_1$ $\pi_2:W_1\oplus W_2 \to W_2$ given by: $\pi_1(w_1,w_2) = w_1$ (some texts write $(w_1,w_2)$ as $w_1+w_2$ or $w_1\oplus w_2$) $\pi_2(w_1,w_2) = w_2$. Let's verify that these are in $\text{Hom}(W_1\oplus W_2,W_1)$ and $\text{Hom}(W_1\oplus W_2,W_2)$, respectively: $\pi_1((w_1,w_2) + (w_1',w_2')) = \pi_1((w_1+w_1',w_2+w_2')) = w_1 + w_1' = \pi_1((w_1,w_2)) + \pi_1((w_1',w_2'))$ so we have an abelian group homomorphism. $\pi_1(g\cdot (w_1,w_2)) = \pi_1((g\cdot w_1,g\cdot w_2)) = g \cdot w_1 = g\cdot \pi_1(w_1,w_2)$, so $\pi_1$ is equivariant. The proof for $\pi_2$ is just the same, I'm sure you can devise your own. Now what we need to come up with is a $G$-module isomorphism: $f:\text{Hom}(V,W_1\oplus W_2) \to \text{Hom}(V,W_1)\oplus \text{Hom}(V,W_2)$. So, given $\eta \in \text{Hom}(V,W_1\oplus W_2)$, define: $f(\eta) = (\pi_1\circ\eta,\pi_2\circ\eta) \stackrel{\text{def}}{\equiv}(\pi_1\circ\eta) + (\pi_2\circ\eta)$ We need to show 2 things: 1. $f$ is a $G$-homomorphism 2. $f$ is a bijection This should get you started, if you have trouble I'll post more later. I believe you'll wind up using the irreducibility of $V$ in showing $f$ is bijective. Thanks from Woita Tags direct, homsets, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shaharhada Abstract Algebra 7 January 23rd, 2014 06:34 PM jrklx250s Real Analysis 3 December 3rd, 2011 03:58 AM MathematicallyObtuse Algebra 4 January 15th, 2011 07:26 PM gianni Abstract Algebra 2 December 2nd, 2010 05:00 AM MathNoob Calculus 4 March 5th, 2009 06:18 AM

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