Problem Dog : Part 1

Algebra Level 5

$x^{932}-1 = 0$ Let $G$ denotes the group of all possible roots of above equation under the operation of multiplication of complex numbers. What is the number of all possible subgroups of the group $G$

The answer is 6.

1 solution

Aman Sharma
May 12, 2018

Let $\omega$ denotes one of the roots of given equation Then, elements of $G$ are $\omega^0=1, \omega,\omega^2,.....,\omega^{931}$ We can see that $G$ is a cyclic group of order $923$ Therefore, converse of Lagrange's Theorem holds Hence, Number of subgroups of $G =$ number of divisors of order of $G =$ number of divisors of $932$ Now, $932=2^2233^1$ Hence, Number of divisors of $932 = (2+1)(1+1) = 3*2 = 6$

Problem Dog : Part 1

The answer is 6.

1 solution

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