My character theory is rather weak, so excuse me if this is a triviality.
I have read on the encyclopedia of maths that for any group $G$, every block of $G$ contains an irreducible character of height zero. Is there a straightforward way to see this? Less importantly, is there a 'canonical' choice, like the trivial character for the principal block ('canonically' can be taken as vaguely as you like)?
Thanks for your time.
Let me make Humphreys' comment just a bit clearer. Let $p^a$ be the full $p$-part of the group order. By definition, the defect of a p-block is the number $d$, minimal such that $p^{a-d}$ divides the degree of every irreducible character in the block. Given an arbitrary irreducible character $\chi$ in a block of defect $d$, therefore, the full $p$-part of $\chi(1)$ is at least $p^{a-d}$, and we can write it as $p^{a - d + h}$, where $h \ge 0$. The integer $h$ is by definition the height of $\chi$. Thus the characters of height zero are exactly the characters in the block whose degrees have the minimum possible $p$-part. Of course, there are always such characters.
Of course, the comment by Jim Humphreys and the answer by Marty Isaacs are entirely accurate. Using the original character-theoretic definition of defect of a block, introduced by Brauer, the existence of a height zero character is almost tautological.
However, there are ring-theoretic definitions of defect of a block which may make the existence of height zero characters less obvious. I illustrate with an example:
If the finite group $G$ has a Sylow $p$-subgroup $P$ of order $p^{a},$ then there is a fairly short ring-theoretic proof by J.A. Green that the dimension of a block $B$ of defect $d$ of $FG$ ($F$ algebraically closed of characteristic $p$) is divisible by $p^{2a-d}$ but by no higher power of $p.$ ( This result had been proved previously by Brauer using characters).
But the dimension of $B$ is also equal to $\sum_{S} {\rm dim}(P(S)) {\rm dim}(S),$
where $S$ runs through the simple $FG$-modules in $B$ and $P(S)$ denotes the projective cover of $S$. Each $P(S)$ has dimension divisible by $p^{a},$ as $P(S)$ is projective,
and each ${\rm dim S}$ is divisible by $p^{a-d},$ (which can be proved in a ring-theoretic fashion via Green's theory of vertices). Thus there is some simple $B$-module $S$ whose dimension is exactly divisible by $p^{a-d}.$ Since Brauer characters of simple $B$-modules are integral combinations of (restrictions to $p$-regular elements of ) ordinary irreducible characters in $B,$ it follows that there is an ordinary irreducible character in $B$ whose degree is divisible by $p^{a-d},$ but by no higher power of $p,$ that is, an irreducible character of height zero.
My purpose is not to point out the shortest proof, merely to point out that it is possible to start from different viewpoints, and still see that there are characters of height zero in every block.
To reinforce further my comment, as well as what Marty and Geoff say, the notion of "defect" of a $p$-block in a group algebra over a sufficiently large field of characteristic $p>0$ was developed by Brauer in a character-theoretic setting. This is presented in the last part of the classic 1962 book by Curtis and Reiner, Representations of Finite Groups and Associative Algebras. This got developed in other styles over the years in work of Green, Alperin, etc., so for instance Alperin's small book Local Representation Theory takes mainly a characteristic $p$ viewpoint without emphasis on ordinary characters. But the notion of height of a character in a $p$-block goes back mainly to Brauer's work and relies on the study of ordinary characters.
Here is a small example which illustrates why the choice of a height 0 character will typically not be canonical (except for blocks of defect 0). Consider the smallest nonabelian simple group, of order 60, realized for example as $\mathrm{PSL}_2(\mathbb{F}_5)$. Here the ordinary irredudible characters have degrees $1, 3, 3, 4, 5$, with the squares of the degrees of course adding up to 60. As in other simple groups of Lie type, there are only two $5$-blocks here. The principal $5$-block has defect 1 (the highest power of 5 in the group order) and involves the first four characters; since $p=5$ divides none of their degrees, all have height 0. But the (Steinberg) character of degreee 5 lies by itself in a $5$-block of defect 0 and has height 0.
