Answer to Question #137506 in Abstract Algebra for Ram

Question #137506
If F is a field with 49 elements, prove that x^49=x. for all x belongs to F. also find characteristic of F.
1
Expert's answer
2020-10-14T18:11:20-0400

Since "F" is a field, "F\\setminus\\{0\\}" is a group with the operation of multiplication ( is zero element of the field "F"). Lagrange's Theorem implies that the order of each element of "F\\setminus\\{0\\}" is a divisor of the order "|F\\setminus\\{0\\}|=48" of "F\\setminus\\{0\\}". Let "x\\in F\\setminus\\{0\\}" be arbitrary. If the order of "x" is equal to "t", then "x^t=e" and "48=ts" for some "s\\in\\mathbb N". Then "x^{48}=x^{ts}=(x^{t})^s=e^s=e". So, "x^{49}=x^{48}x=ex=x" for any "x\\in F\\setminus\\{0\\}". Taking into account that "0^{49}=0", we conclude taht "x^{49}=x" for any "x\\in F." If "p" is a prime number, then the characteristic "char F" of a field "F" of cardinality "p^n" is equal to "p", that is "char F = p." Since "49=7^2", we conclude that "char F = 7."


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