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I found the quoted question in this post interesting: Confusion regarding what kind of isomorphism is intended. I don't have commenting privileges just yet, and since the question already has an accepted answer and it's purpose was not to actually receive an answer to the quoted question, I've decided to post here.

I've been able to do the first part there by using Maschke's theorem and showing that the existence of a two-sided inverse for the projection map from $V$ to a submodule forming part of a decomposition of $V$ leads to a contradiction.

The second part however, I am stumped, could someone kindly provide some guidance?

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    If $\phi$ is an $R$-module endomorphism of $R$ itself, then $\phi(x)=\phi(x\cdot1_R)=x\phi(1_R)$ for all $x\in R$. This allows one to show that $\mathrm{End}_{R\mathrm{-mod}}(R)\cong R^{\,op}$ as rings. Only a bit more is needed beyond this.2012-10-17
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    @anon Thanks a lot for commenting anon, but I'm sorry, I'm not familiar with the notation; what is in the superscript added to $R$ in the last equation?2012-10-17
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    http://en.wikipedia.org/wiki/Opposite_ring2012-10-17
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    @anon I've thought about it, going by your comment that "only a bit more is needed beyond this", I was hoping to have figured it out by now, but am not sure how to use what you have given. Could you please show me how to proceed from what you have given.2012-10-17
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    @HJSprime I think he is saying you can adapt it to your case, where it is not really $End_\mathbb{H}(\mathbb{H})$ but $End_{\mathbb{R}Q}(\mathbb{H})$. There is also a special relationship between $\mathbb{H}$ and $\mathbb{H}^{op}$ which might be part of the hint. Finally, do you care if these are right or left modules? The hint could be elaborated by: $End(R_R)\cong R$ and $End(_RR)\cong R^{op}$ for any ring $R$, using exactly the idea in the first comment.2012-10-17
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    @HJSprime Ah, so I can answer one of my own questions in my comment... in that post I see now the action is on the left.2012-10-17
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    @rschwieb Hi, I've been thinking and searching for a relationship between $\mathbb{H}$ and $\mathbb{H}^{op}$, but have been unable to find one. As for left/right modules, and I am working solely with left modules at the moment, I apologize but I'm not sure what you are hinting at there.2012-10-17
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    @rschwieb Sorry, in case my previous comment didn't make it clear that I would like help from you: How may I adapt it to my case? What is the relationship between $\mathbb{H}$ and $\mathbb{H}^{op}$?2012-10-17
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    @HJSprime I decided it wasn't worth being cryptic about that in my solution, so check it out :)2012-10-17

1 Answers 1

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Hint:

Consider the map $\theta:\mathrm{End}_{\mathbb{R}Q}(\mathbb{H})\rightarrow \mathbb{H^{op}}$ given by $\theta(f)=f(1)$.

Verify that this is a ring homomorphism onto $\mathbb{H^{op}}$. The kernel is therefore an ideal of $\mathrm{End}_{\mathbb{R}Q}(\mathbb{H})$.

I can't justify being cryptic about the fact that $\mathbb{H}^{op}\cong\mathbb{H}$, so I have to give it away :(

Composing that isomoprhism with $\theta$, we now have a homomorphism of $\mathrm{End}_{\mathbb{R}Q}(\mathbb{H})\rightarrow \mathbb{H}$ which is onto. However, the comment at the beginning of the question you linked shows that $\mathrm{End}_{\mathbb{R}Q}(\mathbb{H})$ is a division ring, so... (more hints needed?)

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    Thanks a lot for the answer! So, correct me if I'm wrong, but, first: $\theta$ is onto because, for any quaternion $u$, scalar multiplication on the left by $u^{-1}$ will an endomorphism which $\theta$ maps to $u$. Second, I'm not sure how it is a ring homomorphism considering the law of multiplication in the endomorphism ring is composition? That the kernel is an ideal is simple.2012-10-17
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    In the final bit, the kernel is an ideal, and because the endomorphism ring is a division ring, the ideal is either trivial or the entire ring, but cannot be the entire ring because $\theta$ is of course not identically zero, so the kernel is trivial and so the homomorphism injective and thus an isomorphism.2012-10-17
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    Just need help as to why $\theta$ is a ring homomorphism. Oh, also, I wasn't able to show that the endomorphism ring is in fact a division ring. To do so, using the previous parts of the question, I would need to show that $\mathbb{R}$ is simple as an $\mathbb{R}Q$-module, but I'm not sure how to do this.2012-10-17
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    @HJSprime To verify it's a ring homomorphism, you just need to check additivity and multiplicativity... let me know where the problem is. As I read the post you linked, they were giving you the fact that the endomorphism ring was a division ring in the discussion at the beginning. If you want to do that from scratch too, it'd probably be better to put it in a new question.2012-10-17
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    Sorry, I made a mistake in my first comment, it should be scalar multiplication on the left by $u$.2012-10-17
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    The problem I have is, $\theta(fg) = f \circ g(1) = f(g(1))$, but I need this to equal $f(1)g(1)$.2012-10-17
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    Sorry it's not letting me tag you for some reason :S. Also, in the previous parts, we showed that the endomorphism ring is a division ring if the module is simple, but I'm not sure how to show that $\mathbb{H}$ is a simple $\mathbb{R}Q$-module.2012-10-17
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    You don't have to tag authors of solutions :) For your first problem, use linearity and this trick: $\theta(fg):=f(g(1)\cdot 1)=g(1)\cdot f(1)$. As I read it, the entire solution to your second question is in the first italic paragraph of the post you linked. You do not need its converse (part 1).2012-10-17
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    Oh really, I don't need to tag you? Okay cool. And ah thanks for that, fixes the first issue. For the second issue, in the first italic paragraph it says that "If $V$ is simple, ... , every non-zero element of $\text{End}_{\mathbb{R}G}(V)$ has a two-sided inverse". So, here, I need $\mathbb{H}$ to be simple as an $\mathbb{R}Q$-module, but this I don't yet have.2012-10-17
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    @HJSprime Oh yes, you're right... I don't know how I misread it! Just like in my linearity hint, since the action of $S=\mathbb{R}Q$ matches $\mathbb{H}$ acting on itself, you can find "inverses" in $S$ for elements in the module $\mathbb{H}$. So let $a,b$ be nonzero elements of $\mathbb{H}$. There are "copies" $a',b'$ of $a,b$ and their inverses in $S$. So, $(a'b'^{-1})b=ab^{-1}b=a$. That is, the action of $S$ is transitive on $\mathbb{H}$, i.e. $\mathbb{H}$ is simple.2012-10-17
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    @anon Thanks to both of you, but I'd actually finally been able to figure that out myself.2012-10-17