Can someone give me a simplified proof of Schur's lemma in group theory. Sorry if the question looks a standard textbook proof. But I find the proof complicated in books. It would be helpful if someone can provide a link that proves Schur's lemma in a simpler way.
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$\begingroup$Using and extending notations on the Wiki page.
$\phi: M\to N$ is a linear map and there is a pair of linear representations of the same group $G$: One representation, say $U$, acting on $M$ and the other, say $U$, acting on $N$. The most elementary statement of the lemma established that,
Schur's lemma.
Under the said hypotheses if both the following conditions hold:
(a) $\phi U_g = V_g \phi$ for all $g\in G$
and
(b) $M$ is irreducible for $U$, $N$ is irreducible for $V$,
then $\phi$ is bijective or it is the null map.
Proof. Suppose that $\phi$ is not bijective, our final goal is obviously to prove that $\phi$ is the null map.
If $\phi$ is not bijective, it may happen for two reasons only:
(1) $Ran(\phi)$ is smaller than $N$ or
(2) $Ker(\phi)$ is larger than $\{0\} \subset M$ (or both).
Let us examine separately these cases.
Suppose (1) is valid. $y\in Ran(\phi)$ is equivalent to say that $y= \phi x$ for some $x\in M$. For such $y$, in view of (a) we have $V_gy = V_g\phi x = \phi U_g x = \phi U_gx' \in Ran(\phi)$. All that entails $$V_g Ran(\phi) \subset Ran(\phi)\quad \mbox{for all $g \in G$.}$$ As $V$ is irreducible for (b), it must be (i) $Ran(\phi)=\{0\}$ or (ii) $Ran(\phi)= N$. In the case (i) $\phi$ is the null map and the proof concludes, in the case (ii) case, since $\phi$ is not bijective by hypotheses,we must have that $Ker(\phi)$ is larger than $\{0\}$ and so we can restrict to the other case treated below.
Suppose (2) is valid. In this case (a) implies: $$U_g Ker(\phi) \subset Ker(\phi)\quad \mbox{for all $g \in G$.}$$
Indeed: If $x \in Ker(\phi)$ then $\phi x=0$ and so, by (a), $\phi U_gx =V_g\phi x = V_g 0 =0$, so $U_gx \in Ker(\phi)$ as wanted.
Since $U$ is irreducible for (b), it must be $Ker(\phi)=M$ and this means that $\phi$ is the null map and the proof stops, or $Ker(\phi)=\{0\}$ that is impossible by hypotheses ($Ker(\phi)$ is larger than $\{0\}$).
In all cases $\phi$ is the null map if it is not bijective. QED
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