Abstract Algebra help wanted :(

[Rebecca123]

Guys, can anybody help?

Let Q [√ 5]:= {a+b√5; a,b ϵ Q)

Show that (Q[√ 5], +) is a subgroup of (C,+), and that it is generated by two elements.
Show that (Q[√ 5]-{0}), .) is a subgroup of (C,.)
Q being rational, C being complex and the little dot being multiply.

Cheers guys :wink:

 

[pka]

Guys, can anybody help?

Let Q [√ 5]:= {a+b√5; a,b ϵ Q)

Show that (Q[√ 5], +) is a subgroup of (C,+), and that it is generated by two elements.
Show that (Q[√ 5]-{0}), .) is a subgroup of (C,.)
Q being rational, C being complex and the little dot being multiply.

You must show that both of those sets are closed with respect to operation and inverse.

 

[Ishuda]

Guys, can anybody help?

Let Q [√ 5]:= {a+b√5; a,b ϵ Q)

Show that (Q[√ 5], +) is a subgroup of (C,+), and that it is generated by two elements.
Show that (Q[√ 5]-{0}), .) is a subgroup of (C,.)
Q being rational, C being complex and the little dot being multiply.

Cheers guys :wink:

Go through the definition: G is a group under • if it satisfies
Closure
For all a, b in G, the result of the operation, a • b, is also in G.
Associativity
For all a, b and c in G, (a • b) • c = a • (b • c).
Identity element
There exists an element e in G, such that for every element a in G, the equation e • a = a • e = a holds. Such an element is unique, and thus one speaks of the identity element.
Inverse element
For each a in G, there exists an element b in G such that a • b = b • a = e, where e is the identity element.

 

[Rebecca123]

Thanks you guys

 

[HallsofIvy]

Go through the definition: G is a group under • if it satisfies
Closure
For all a, b in G, the result of the operation, a • b, is also in G.
Associativity
For all a, b and c in G, (a • b) • c = a • (b • c).
Identity element
There exists an element e in G, such that for every element a in G, the equation e • a = a • e = a holds. Such an element is unique, and thus one speaks of the identity element.
Inverse element
For each a in G, there exists an element b in G such that a • b = b • a = e, where e is the identity element.

But here, your set is a subset of C and you are only asked to prove it is a subgroup of C. You already know (I hope!) that that C is a group under + and C- {0} is a group under \(\displaystyle \cdot\) so you already know it is associative, has an identity, and the operation is associative. You really only need to show that this subset is closed under the operation and closed under taking the inverse, as pka said.
(And, technically, show that it is non-empty but 0 is obviously in this set.)