Hello As I checked the atlas of finite simple groups if $p$ is a prime then according to the notations of the atlas of finite simple groups, $2.L_2(p)$ has an irreducible character of degree $(p-1)/2$ or $(p+1)/2$. Why this character exists and is it true for each prime number $p>3$? Thanks so much for your help.
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1$\begingroup$ I can tell you that $L_2(p)$ always has an irreducible character of degree $p$ for any prime $p$. Recall that $L_2(p)$ is simply $G/Z(G)$ where $G$ is $SL_2(p)$. The character of $L_2(p)$ of degree $p$ is then simply obtained from the Steinberg character of $SL_2(p)$. This is defined, for instance, in Digne and Michel's book "Representations of Finite Groups of Lie Type". Unfortunately I don't know enough about double covers to tell you why you get the character of degree 2p. However it should be a character covering the Steinberg character. $\endgroup$– Jay TaylorCommented Mar 2, 2013 at 9:32
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$\begingroup$ Thanks, Jay, for explaining that $L_2(p)$ is just $PSL_2(p)$. But what is $2.L_2(p)$ exactly? $\endgroup$– Alain ValetteCommented Mar 2, 2013 at 11:36
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1$\begingroup$ @Alain: The notation ${\rm L}_n(q)$ for ${\rm PSL}(n,q)$ is often used by people working on finite simple groups, in the spirit that series of simple groups deserve simple (i.e. one-letter) names. $\endgroup$– Stefan Kohl ♦Commented Mar 2, 2013 at 11:42
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$\begingroup$ I am sorry for the mistake in the question. In fact I want to ask that is it true that the non-split extension of $L_2(p)$ by $Z_2$ has an irreducible character of degree $(p-1)/2$ or $(p+1)/2$ in general. In fact in the atlas of finite group we see that $2.L_2(11)$ has an irreducible character of degree 6, or $2.L_2(13)$ has an irreducible character of degree 8. $\endgroup$– BHZCommented Mar 2, 2013 at 11:45
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$\begingroup$ @Stefan: you mean that $2.L_2(P)$ is $SL_2(p)$? Oh boy... $\endgroup$– Alain ValetteCommented Mar 2, 2013 at 11:54
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OK, so if I am not mistaken the non-split extension of $L_2(p)$ should simply be $SL_2(p)$. You are now asking whether $SL_2(p)$ has a character of degree $(p-1)/2$ or $(p+1)2$ when $p \neq 2$. Indeed this is true. The generic character table of $SL_2(q)$ is given in Table 5.4 of Bonnafe's book "Representations of $SL_2(q)$". We see from this table that the degrees of the irreducible characters of $SL_2(p)$ are contained in the list
$$1,p,\frac{p+1}{2},\frac{p-1}{2},p+1,p-1$$
There is only one character of degree 1, the trivial character, and one character of degree $p$, the Steinberg character. There are then two characters of degree $\frac{p+1}{2}$ and two characters of degree $\frac{p-1}{2}$. These occur as irreducible constituents of the restriction of an irreducible Deligne-Lusztig character of $GL_2(p)$. When $p=2$ then one obtains the character table of $SL_2(p)$ simply by restricting the characters of $GL_2(p)$.
You may also find this useful by Mark Reeder.
answered Mar 2, 2013 at 10:48
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$\begingroup$ Sorry, I was in the process of changing an answer to the OP's original question before he changed it. The above now explains why $SL_2(p)$ has irreducible characters of degree $\frac{p-1}{2}$ and $\frac{p+1}{2}$. $\endgroup$ Commented Mar 2, 2013 at 13:02