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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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    @HJSprime I decided it wasn't worth being cryptic about that in my solution, so check it out :)2012-10-17

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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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    @anon Thanks to both of you, but I'd actually finally been able to figure that out myself.2012-10-17