2 HW) 1st option true. (x³+1 has root -1) 2nd option is wrong as x²+1 has no root in R. 3rd option is true as C is algebraically closed and if root belongs to Q' nothing to do, if root belongs to Q then as polynomial is monic that root belongs to Z and hence belongs to the given union. 4th option is wrong as root is member of Q implies it is a member of Z. So Q\Z contains no roots of monic polynomial over Z.
1HW) 1st option is wrong as sub group of cyclic group is cyclic and G is not cyclic. 3rd option is wrong as product of abelian groups is abelian and G is not abelian. 4th option is wrong as every element of Q has infinite order. 2nd option is true as we can define a map from G to Z2 by mapping the cycles to their signature which is onto.
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First question - b)
Second question -ac)
2 HW) 1st option true. (x³+1 has root -1)
2nd option is wrong as x²+1 has no root in R.
3rd option is true as C is algebraically closed and if root belongs to Q' nothing to do, if root belongs to Q then as polynomial is monic that root belongs to Z and hence belongs to the given union.
4th option is wrong as root is member of Q implies it is a member of Z. So Q\Z contains no roots of monic polynomial over Z.
1st HW: (b)
2nd HW:(a)
1HW) 1st option is wrong as sub group of cyclic group is cyclic and G is not cyclic.
3rd option is wrong as product of abelian groups is abelian and G is not abelian.
4th option is wrong as every element of Q has infinite order.
2nd option is true as we can define a map from G to Z2 by mapping the cycles to their signature which is onto.