1. ## Groups prob. Help needed

1)consider Z24 as additive group modulo 24.Then find the number of element of order 8 in group 24 ?

2)Consider Z5 and Z20 as ring modulo 5 and 20.Them the number of homomorphisms
p:Z5-->Z20 is?

3)if Q is a field of rational numbers and Z2 as a field of modulo 2 let
f(x)=x^3-9x^2+9x+3 then witch of following is correct.1)f(x) is irreducible over Z2
2)f(x) is irreducible over both Q and Z2
3)f(x) is reducible over Q but irreducible and Z2
4)f(x) is reducible over both Z2 and Q

(these questions asked by indian institue of technology delhi asked in gate test )

2. Originally Posted by reflection_009
1)consider Z24 as additive group modulo 24.Then find the number of element of order 8 in group 24 ?
this is asking for the number of solutions to $\displaystyle [8a]_{24} = [0]_{24}$. that is, how many solutions are there to the congruence equation $\displaystyle 8a \equiv 0~(\text{mod }24)$

3. Originally Posted by reflection_009
1)

3)if Q is a field of rational numbers and Z2 as a field of modulo 2 let
f(x)=x^3-9x^2+9x+3 then witch of following is correct.1)f(x) is irreducible over Z2
2)f(x) is irreducible over both Q and Z2
3)f(x) is reducible over Q but irreducible and Z2
4)f(x) is reducible over both Z2 and Q
Since f is of degree one, f is reducible if and only if f has a linear factor.
Now recall that x-c is a linear factor if f(c)=0. So basically we only need to check if there is an element c in the field that make f(c)=0
for Z2, we have f(0),f(1) not equal zeero so f is irreducible there. For Q, by Gauss Lemma we need only to check if f has an integer root. This integer root must divides the constant term, i.e. 3. So you need to check f(1),f(-1),f(3),f(-3). If one of them is zero then it is reducible but if not then it is irreducible.

4. Originally Posted by reflection_009
1)consider Z24 as additive group modulo 24.Then find the number of element of order 8 in group Z24 ?
Let $\displaystyle 3\mathbb{Z}_{24} = \left< [3] \right>$. This is a subgroup of order $\displaystyle 24/3 = 8$. Any element of order 8 must lie in this subgroup. Since $\displaystyle 3$ is a generator it means all other generators i.e. elements of order 8 are $\displaystyle k[3]$ where $\displaystyle \gcd(k,8)=1$. Therefore the are $\displaystyle \phi(8)$ such elements.

2)Consider Z5 and Z20 as ring modulo 5 and 20.Them the number of homomorphisms
p:Z5-->Z20 is?
Hint: If $\displaystyle \phi$ is a homomorphism then it is completely determined by its value on $\displaystyle \phi ([1]_5])$. Find the possibilities that this can be.

3)if Q is a field of rational numbers and Z2 as a field of modulo 2 let
f(x)=x^3-9x^2+9x+3 then witch of following is correct.1)f(x) is irreducible over Z2
2)f(x) is irreducible over both Q and Z2
3)f(x) is reducible over Q but irreducible and Z2
4)f(x) is reducible over both Z2 and Q
$\displaystyle x^3 - 9x^2+9x+3$ is irreducible over $\displaystyle \mathbb{Z}_2$ if and only if it has no zeros in $\displaystyle \mathbb{Z}_2$. Likewise for $\displaystyle \mathbb{Q}$.

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# number of ring homomorphism z5 to z20

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