On existence of a certain irreducible character of $SL(5, q)$ Let $q=p^f$ be a prime power such that $q \equiv 1 \pmod 5$. According to the list of irreducible (complex) character degrees of $SL(5, q)$ in Frank Luebeck's homepage (here), $SL(5, q)$ has 20 irreducible characters of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$. Using Lusztig's parametrization of irreducible complex characters of groups of Lie type, I want to prove the existence of such an irreducible character for $SL(5, q)$ when $q$ is sufficiently large: Assume that $\chi \in \mathrm{Irr}(SL(5, q))$ is of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$. By Lusztig's result, $\chi$ must belong to a Lusztig's rational serie $\mathcal{E}(SL(5, q), s)$, where $s$ is a semisimple element of $PGL(5, q)$. Moreover, the degree of $\chi$ is given by $\chi(1)=\frac{|SL(5, q)|_{p'}}{|C_{PGL(5, q)}(s)|_{p'}}\psi_s(\chi)(1)$, where $\psi_s(\chi)(1)$ is the degree of a unipotent character of $C_{PGL(5, q)}(s)$ (see Theorem 13.23 and Remark 13.24 of Digne and Michel's book). It is known that if $p$ is sufficiently large (I think $p>5$ would be sufficient), then all the $p'$-degree irreducible characters of $SL(5, q)$ are preciesly the so-called semisimple characters. Therefore, $\chi=\chi_s$ is a semisimple character and $\chi(1)=\frac{|SL(5, q)|_{p'}}{|C_{PGL(5, q)}(s)|_{p'}}$. Thus, the existence of such $\chi$ depends on the existence of a semisimple element $s \in PGL(5, q)$ such that $|C_{PGL(5, q)}(s)|_{p'}=5(1+q+q^2+q^3+q^4)$. Question: Is there any semisimple element $s$ of $PGL(5, q)$ satisfying $|C_{PGL(5, q)}(s)|_{p'}=5(1+q+q^2+q^3+q^4)$? Some Thoughts: We deduce from above argument that we must have $|C_{PGL(5, q)}(s)|_{p'}=5(1+q+q^2+q^3+q^4)$. So if $s_1$ is a preimage of $s$ in $GL(5, q)$, the characteristic polynomial of $s_1$ is an irreducible polynomial of degree $5$ in $\mathbb{F}_q[x]$. Hence we have $C_{GL(5, q)}(s_1)\cong GL(1, q^5)$ and $C_{PGL(5, q)}(s)\cong C_{GL(5, q)}(s_1)/Z(GL(5, q))$. But this implies that $|C_{PGL(5, q)}(s)|_{p'}=1+q+q^2+q^3+q^4$ and so $SL(5, q)$ could not contain an irreducible character of degree$1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$! How can I explain this? I would be grateful if you could hint me how to find a proper semisimple element corresponds to $\chi$.
 A: [The author or this question made me aware of this thread, so I send the answer here.]
The description in the question is almost correct, except when it comes to the centralizer of the semisimple element $s$ in $PGL_5(q)$. For $q \equiv 1 \pmod 5$ the characters of $SL_5(q)$ of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q−1)^4$  are in Lusztig series corresponding to certain elements $s$ of order $5$ in the dual group $PGL_5(q)$. Preimages $s_1 \in GL_5(q)$ of such $s$ can be found as elements in the cyclic Coxeter torus (a maximal torus of order $q^5-1$) which have an eigenvalue $x$ of order $5(q-1)$. Its other eigenvalues are $x^q, x^{q^2}, x^{q^3}, x^{q^4}$ (the successive quotients of these eigenvalues are a fixed $5$-th root of unity). 
Such  $s_1 \in GL_5(q)$ are regular and have a torus of order $q^5-1$ as centralizer. Its image $s \in PGL_5(q)$ has a non-connected centralizer generated by a maximal torus and a $5$-cycle in the Weyl group. 
To $s_1$ corresponds one irreducible character of $GL_5(q)$ (which is isomorphic to its dual group) and to $s$ corresponds the restriction of this character to $SL_5(q)$ which splits into $5$ irreducible constituents of same degree, these are the characters in question. Since there are $4$ primitive roots of $5$ we get altogether $4 \cdot 5 = 20$ characters of that degree.
