@HelenS: I see your problem. Perhaps you could settle with the following solution? Assign your integral-expression to expr, say, and then use op(3,expr) to print the nontrivial part of the calculation. In this way, the output renders nicely with respect to line breaking.

@HelenS: I had a hunch that that was indeed the problem. Can you provide an example? Either a code snippet producing such output, or the output itself uploaded (using the green arrow in the text editor) as an image perhaps?

@Mac Dude: The :- does not go away if I insert a 'restart' at the beginning of the worksheet. I am using Maple 17.

@MrYouMath: Thanks for your thumbs up on my answer, although I think I have not deserved it :-).

@ecterrab: Thanks for correcting my mistake. Having had zero experience with determining symmetries of differential equations using Maple, I was not really sure myself of the soundness of my proposal. Admittedly, it was thus somewhat a long shot.

PS: My thumbs up for your answer above.

@mskalsi: Thanks. Now I can indeed verify the transformation, and thus the isomorphism.

PS: Concerning your original question as to how to have Maple check for such isomorphisms, I am still largely in the dark. I am still surprised that their seems to be no built-in/fundamental command in the LieAlgebras package for doing such a thing.

@vv: My apology. Obviously, I did not read all of your post properly. Your f[%] does give the matching elements.

@mskalsi: I do not understand your transformation. How can e1 be mapped to both -e3 and e2, and how can e1 and e3 be mapped to the same element e2?

@vv: You select construction returns the indices of matching elements. That is not, I think, what the poster had in mind, writing f[i] rather than i. I thought about using the following construction instead:

zip((x,y) -> `if`(x = y,x,NULL),f,h);

@Carl Love: No, that is not correct. I tried it myself, and found that it produced {1,2,4} from which I concluded that that was probably not what the poster had in mind, saying that the output should be {1,2}. But that, of course, may depend on the correctly raised objection by VV, i.e., whether sets or lists are used.

@mskalsi: I am beginning to think that I am mistaken, that my 'solution' above is erroneous, the reason being that I seem unable to make sense of the transformation between Lie algebra bases provided by an invertible solution A. The transformed basis does not (readily, if at all) satisfy the new Lie algebra. Perhaps others can weigh in with a solution, or with good ideas of how to proceed.

Update I: The origin of my error seems to be the following: A homomorphism h of a Lie algebra is a map for which

[h(g[a]),h(g[b])] = f[a,b,c]*h(g[c]),

where g[a] are the basis elements (the generators of the corresponding Lie group), and where f[a,b,c] are the structure constants, sum of c being implicit. Thus it has nothing a priori to do with the structure constants f'[a,b,c], note prime, of some other Lie algebra (the algebra alg2, say). But then I do not understand the meaning of the Query as given in the mentioned Example 2.

Update II: A low-level approach to checking for isomorphism between two Lie algebras would be to find, if possible, an invertible matrix A for which

A[a,d]*A[b,e]*f[d,e,f] = A[c,f]*f'[a,b,c],

summation over c,d,e, where f[a,b,c] and f'[a,b,c], note prime, are the structure constants of the two Lie algebras.

@mmcdara: Thanks for clearing that up for me. Your guess seems entirely plausible.

@sand15: Is your expression for S2 correct? The map returns a list of lists, each of the latter ones containing two copies of the same integer.

@Carl Love: As I suspected, I had missed the point :-).

Thanks. Now it executes.