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Let $K$ be a field, $G$ group, $V$ and $W$ be finite dimensional $K[G]$ modules and $X$ and $Y$ be the representations afforded by $V$ and $W$ respectively.I need to show that $X$ and $Y$ are equivalent if and only when $V$ and $W$ are as $K[G]$ modules isomorphic . Any hints !

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What is your definition of equivalent? Giving the same homomorphism of $G$ into a matrix group up to conjugation? – Brett Frankel Apr 17 '12 at 21:06
yes thats what i mean. X=ZYZ^-1 for some Z . – Theorem Apr 17 '12 at 21:09
up vote 1 down vote accepted

Let $\varphi$ be an isomorphism from $V$ to $W$. Look at the action of $g$ on $W$ by writing it in terms of $\varphi$ and action on $V$. (Remember, the action of a ring on a module commutes with module homomorphisms). That should hopefully be enough of a hint. Let me know if you're still confused after trying to work this out.

EDIT: $V=\varphi(\varphi^{-1} (W))$. So if $v \in V$, $g\cdot v=g\cdot \varphi(\varphi^{-1} (v))=\varphi(g\cdot\varphi^{-1} (v))$ In other words, acting by $g$ is the same as mapping $V$ to $W$, acting by $g$ in $W$, and then mapping back to $V$. Writing $\varphi$ as the linear map $Z$, we get that for any $g\in G$, $X(g)=Z\cdot Y(g)\cdot Z^{-1}$

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I am not able to see much further from what you said . I just applied the isomorphism between V and W and wrote group elements in terms of 'Phi'. after that i am stuck :( – Theorem Apr 17 '12 at 21:28
Let me know if it makes sense now. I think this spells things out in more or less full detail. – Brett Frankel Apr 17 '12 at 21:45
Thanks :) one more thing how did you move 'g' inside ? – Theorem Apr 17 '12 at 21:55
By definition, ring action commutes with module homomorphisms. – Brett Frankel Apr 17 '12 at 22:02

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