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For a class I'm taking this semester, I was given this question:

(To guarantee clarity, I am quoting the full question even though my own question is only regarding the final sentence.)

Let $G$ be a finite group. If $V$ is an $\mathbb{R}G$-module, then $\text{End}_{\mathbb{R}G}(V)$ denotes the ring of $\mathbb{R}G$-module endomorphisms of $V$ (where the multiplication is composition). If $V$ is simple, then by the same argument as in the proof of Schur's Lemma part (1), every nonzero element of $\text{End}_{\mathbb{R}G}(V)$ has a two-sided inverse.

(i) Prove that the converse is true: if every non-zero element of $\text{End}_{\mathbb{R}G}(V)$ has a two-sided inverse, then $V$ is simple.

(ii) Let $\mathbb{H}$ be the quaternion ring, $Q$ the quaternion group $\{\pm 1,\pm i, \pm j, \pm k\}$ and $T : Q \rightarrow GL(\mathbb{H})$ the linear representation of $Q$ defined by $T(g)(v) = gv$. Regard $\mathbb{H}$ as an $\mathbb{R}Q$-module via the representation $T$. Show that $\text{End}_{\mathbb{R}Q}(\mathbb{H}) \cong \mathbb{H}$.

Now I have completed part (i), am attempting to do part (ii). My question is, $\text{End}_{\mathbb{R}Q}(\mathbb{H})$ denotes a ring, while $\mathbb{H}$ we are regarding as an $\mathbb{R}Q$-module, so exactly what kind of isomorphism are we to show exists?

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In the expression $\mathrm{End}_{\mathbb R Q}(\mathbb H)$, you are regarding $\mathbb H$ as an $\mathbb R Q$-module, but in the isomorphism $\mathrm{End}_{\mathbb R Q}(\mathbb H) \cong \mathbb H$, you are supposed to once more regard $\mathbb H$ as a ring (with its usual ring structure).

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    @MattE Ok, good to know!2012-10-17