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Sep 19, 2018 at 10:48 comment added Geoff Robinson Note that Dixon's argument can be simplified a little. No finite simple group can have an irreducible complex character of degree $2$. If $G$ were a simple group with such a representation, then $|G|$ would be even (since irreducible character degrees divide the group order). Then $G$ would contain an involution $t$ by Cauchy's theorem, and $t$ would be in $Z(G)$ since $t$ would be represented by $-I$ (on consideration of determinant), contrary to simplicity.
Sep 19, 2018 at 6:58 history edited YCor CC BY-SA 4.0
Added reference info, will help in case link is broken.
Sep 19, 2018 at 3:45 history answered Keith Kearnes CC BY-SA 4.0