I am a student in computer science - first year. I study linear linear algebra 2 - course of linear algebra 1. - In some institutions academic studies teach the courses together / teach in another way.
I solved the question - but the way it is rather strange - without permutations.
"Consider the groups ({e, a, b, c}, ·) – that is, the groups of 4 elements. There are precisely two such non-isomorphic groups (that is, essentially different). Here are the partially filled multiplication tables for them:
1) $$ \begin{array}{c|lcr}*&e&a&b&c\\ \hline e&e&a&b&c\\a&a&e&\\b&b&&\\c&c\end{array}$$ 2) $$ \begin{array}{c|lcr}*&e&a&b&c\\ \hline e&e&a&b&c\\a&a&b&\\b&b&&\\c&c\end{array}$$
Fill in the contents, so that the two will be non-isomorphic groups. "
i fill the table and got :
1) $$ \begin{array}{c|lcr}*&e&a&b&c\\ \hline e&e&a&b&c\\a&a&e&c&b\\b&b&c&a&e\\c&c&b&e&a\end{array}$$
I have rejected the possibilities that I believe do not exist - for example: bb = a - other option impossible. bb cannot be e - lagrange's theorem, or c/b - because i can get 2 same element in the same row/column.
2) $$ \begin{array}{c|lcr}*&e&a&b&c\\ \hline e&e&a&b&c\\a&a&b&c&e\\b&b&c&e&a\\c&c&e&a&b\end{array}$$
fill like a sudoko.
I did not solve algebraic way - I think - is there a different way to solve these questions?
Hint: The other, partially filled, is
$$ \begin{array}{c|lcr}*&e&a&b&c\\ \hline e&e&a&b&c\\a& a & e & & \\b&b& &e& \\c&c& & &e\end{array}.$$
The sudoku-like nature is because the tables are what are known as Latin squares.
Another way to show that there is only two is to use the fundamental theorem of abelian groups, although that is overkill.
Yet another way is to assume such a group is not (isomorphic to) one of them, only to prove that it's (isomorphic to) the other, since $A\lor B\equiv (\lnot A)\to B$ for any statements $A, B$.