TITLE: Real representation of finite groups 
QUESTION [6 upvotes]: I want to know if there is complete theory on real representation of finite
groups.
Say, given a finite group G, can we know all the injections from G to 
GL(n,R) for a particular n?

REPLY [3 votes]: Some of the other answers omit several details and I had already written this on another forum.  At any rate, perhaps the example would be useful:
More or less yes, though for larger groups it is not very practical.
Calculate the character table of G. Use the Frobenius-Schur indicator to
fix the complex and quaternionic irreducible characters to be characters of
irreducible real representations. Take all non-negative integer combinations
of these characters of real reps whose total degree is n and whose kernel is
trivial (the degree and the kernel of a character are very easy to calculate).
There is a 1-1 correspondence between injections of G into GL(n,R) up to GL(n,R)
conjugacy and characters of real reps whose total degree is n and whose kernel
is trivial.
Except for cyclic groups G and n=1, there will be uncountable many injections
into GL(n,R), so you'll want to only consider them up to GL(n,R) conjugacy in
order to get a finite number.
Also notice that if f1:G→GL(m1,R) is an injection and f2:G→GL(m2,R) is any group
homomorphism, then another injection is "f1⊕f2":G→GL(m1+m2,R):g→[f1(g),0;0,f2(g)],
the function that takes the two matrices from f1 and f2 and makes them the blocks
of a block diagonal matrix. For this reason the number of injections, even up to
conjugacy, can get unwieldy.

Example: The non-cyclic group of order four:
For G the Klein four-group the irreducible characters are all real, and are
[1,1,1,1], [1,-1,-1,1], [1,-1,1,-1,], [1,1,-1,-1]. The degree of each of these
is 1, the first number. The kernel of each of these are the positions where
the value and the degree are equal, so all of them have a kernel. If n was 10,
then I'd look at all the ways of choosing non-negative integers a,b,c,d such
that a+b+c+d=10 (so that the degree was 10) and such that at least 2 of b,c,d
were non-zero (so that the kernel was trivial). There are 282 such combinations,
one of which is a=5, b=0, c=2, d=3. This corresponds to the character
5*[1,1,1,1]+0*[1,-1,-1,1]+2*[1,-1,1,-1]+3*[1,1,-1,-1] = [ 10, 6, 4, 0 ]. It has
degree 10 (the first and largest positive number) and trivial kernel (all the
other positive numbers are strictly smaller). It corresponds to the representation
g1 → diag( 1,1,1,1,1, -1,-1, 1,1,1 ) and g2 → diag( 1,1,1,1,1, 1,1, -1,-1,-1 )
and of course g1*g2 → diag( 1,1,1,1,1, -1,-1, -1,-1,-1 ). This is 5 copies of
the representation g→1, g2→1, followed by two copies of g1→-1, g2→1, and
finally three copies of g1→1, g2→-1. Since b=0 we did not include any copies
of g1→-1, g2→-1.
By some silly counting, the number of injections of the Klein four-group into
GL(n,R) up to GL(n,R) conjugacy is (n^3+6*n^2+11*n-18)/6, so it grows somewhat
quickly but not too crazily.

For non-abelian groups G, the conversion of the characters into the matrices
is a little harder (for abelian groups, the matrices are all 1x1 and really
are just the character values), but at some point the matrices are too big
anyways, and the characters are a more efficient data type.