must appear in the GROUP BY clause or be used in an aggregate function

Yes, this is a common aggregation problem. Before SQL3 (1999), the selected fields must appear in the GROUP BY clause[*].

To workaround this issue, you must calculate the aggregate in a sub-query and then join it with itself to get the additional columns you'd need to show:

SELECT m.cname, m.wmname, t.mx
FROM (
    SELECT cname, MAX(avg) AS mx
    FROM makerar
    GROUP BY cname
    ) t JOIN makerar m ON m.cname = t.cname AND t.mx = m.avg
;

 cname  | wmname |          mx           
--------+--------+------------------------
 canada | zoro   |     2.0000000000000000
 spain  | usopp  |     5.0000000000000000

But you may also use window functions, which looks simpler:

SELECT cname, wmname, MAX(avg) OVER (PARTITION BY cname) AS mx
FROM makerar
;

The only thing with this method is that it will show all records (window functions do not group). But it will show the correct (i.e. maxed at cname level) MAX for the country in each row, so it's up to you:

 cname  | wmname |          mx           
--------+--------+------------------------
 canada | zoro   |     2.0000000000000000
 spain  | luffy  |     5.0000000000000000
 spain  | usopp  |     5.0000000000000000

The solution, arguably less elegant, to show the only (cname, wmname) tuples matching the max value, is:

SELECT DISTINCT /* distinct here matters, because maybe there are various tuples for the same max value */
    m.cname, m.wmname, t.avg AS mx
FROM (
    SELECT cname, wmname, avg, ROW_NUMBER() OVER (PARTITION BY avg DESC) AS rn 
    FROM makerar
) t JOIN makerar m ON m.cname = t.cname AND m.wmname = t.wmname AND t.rn = 1
;


 cname  | wmname |          mx           
--------+--------+------------------------
 canada | zoro   |     2.0000000000000000
 spain  | usopp  |     5.0000000000000000

[*]: Interestingly enough, even though the spec sort of allows to select non-grouped fields, major engines seem to not really like it. Oracle and SQLServer just don't allow this at all. Mysql used to allow it by default, but now since 5.7 the administrator needs to enable this option (ONLY_FULL_GROUP_BY) manually in the server configuration for this feature to be supported...

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In Postgres, you can also use the special DISTINCT ON (expression) syntax:

SELECT DISTINCT ON (cname) 
    cname, wmname, avg
FROM 
    makerar 
ORDER BY 
    cname, avg DESC ;

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The problem with specifying non-grouped and non-aggregate fields in group by selects is that engine has no way of knowing which record's field it should return in this case. Is it first? Is it last? There is usually no record that naturally corresponds to aggregated result (min and max are exceptions).

However, there is a workaround: make the required field aggregated as well. In postgres, this should work:

SELECT cname, (array_agg(wmname ORDER BY avg DESC))[1], MAX(avg)
FROM makerar GROUP BY cname;

Note that this creates an array of all wnames, ordered by avg, and returns the first element (arrays in postgres are 1-based).

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For me, it is not about a "common aggregation problem", but just about an incorrect SQL query. The single correct answer for "select the maximum avg for each cname..." is

SELECT cname, MAX(avg) FROM makerar GROUP BY cname;

The result will be:

 cname  |      MAX(avg)
--------+---------------------
 canada | 2.0000000000000000
 spain  | 5.0000000000000000

This result in general answers the question "What is the best result for each group?". We see that the best result for spain is 5 and for canada the best result is 2. It is true, and there is no error. If we need to display wmname also, we have to answer the question: "What is the RULE to choose wmname from resulting set?" Let's change the input data a bit to clarify the mistake:

  cname | wmname |        avg           
--------+--------+-----------------------
 spain  | zoro   |  1.0000000000000000
 spain  | luffy  |  5.0000000000000000
 spain  | usopp  |  5.0000000000000000

Which result do you expect on runnig this query: SELECT cname, wmname, MAX(avg) FROM makerar GROUP BY cname;? Should it be spain+luffy or spain+usopp? Why? It is not determined in the query how to choose "better" wmname if several are suitable, so the result is also not determined. That's why SQL interpreter returns an error - the query is not correct.

In the other word, there is no correct answer to the question "Who is the best in spain group?". Luffy is not better than usopp, because usopp has the same "score".

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SELECT t1.cname, t1.wmname, t2.max
FROM makerar t1 JOIN (
    SELECT cname, MAX(avg) max
    FROM makerar
    GROUP BY cname ) t2
ON t1.cname = t2.cname AND t1.avg = t2.max;

Using rank() window function:

SELECT cname, wmname, avg
FROM (
    SELECT cname, wmname, avg, rank() 
    OVER (PARTITION BY cname ORDER BY avg DESC)
    FROM makerar) t
WHERE rank = 1;

Note

Either one will preserve multiple max values per group. If you want only single record per group even if there is more than one record with avg equal to max you should check @ypercube's answer.

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This seems to work as well

SELECT *
FROM makerar m1
WHERE m1.avg = (SELECT MAX(avg)
                FROM makerar m2
                WHERE m1.cname = m2.cname
               )

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I recently run into this problem, when trying to count using case when, and found that changing the order of the which and count statements fixes the problem:

SELECT date(dateday) as pick_day,
COUNT(CASE WHEN (apples = 'TRUE' OR oranges 'TRUE') THEN fruit END)  AS fruit_counter

FROM pickings

GROUP BY 1

Instead of using - in the latter, where I got errors that apples and oranges should appear in aggregate functions

CASE WHEN ((apples = 'TRUE' OR oranges 'TRUE') THEN COUNT(*) END) END AS fruit_counter

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