prove problem

[mathsfun]

can somebody help?

 

prove or disprove, If G is a group in which every proper subgroup is cyclic, then G is cyclic.

 

in additon, can u explain cyclic

[mathsfun]

can somebody help?

 

prove or disprove, If G is a group in which every proper subgroup is cyclic, then G is cyclic.

 

in additon, can u explain cyclic

[bloodhound]

What is a "proper" subgroup? never came across that term before.

 

If G is a group, and you have a [math]g \in G[/math] such that [math]G=<g>[/math], then G is called a cyclic group and g is called the generator of G

 

Here,

[math]<g>=\{g^k:k \in \mathbb{Z}\}[/math]

or additively

[math]<g>=\{mg:m \in \mathbb{Z}\}[/math]

 

Well basically its just saying that, if u have a group G, then its a cyclic group if there exists an element in G such that if u take all the integer powers of that element, you get the group G.

 

A nice property is that a subgroup of a cyclic group is also cyclic.

 

Cyclic groups are also abelian by the way.

[bloodhound]

What is a "proper" subgroup? never came across that term before.

 

If G is a group, and you have a [math]g \in G[/math] such that [math]G=<g>[/math], then G is called a cyclic group and g is called the generator of G

 

Here,

[math]<g>=\{g^k:k \in \mathbb{Z}\}[/math]

or additively

[math]<g>=\{mg:m \in \mathbb{Z}\}[/math]

 

Well basically its just saying that, if u have a group G, then its a cyclic group if there exists an element in G such that if u take all the integer powers of that element, you get the group G.

 

A nice property is that a subgroup of a cyclic group is also cyclic.

 

Cyclic groups are also abelian by the way.

[Dogtanian]

At a guess bloodhound, I say a "proper" subgroup was just a subgroup of G that in't G itself, saying that though, I don't think I've conme across the term before either...*thinks*...

[Dogtanian]

At a guess bloodhound, I say a "proper" subgroup was just a subgroup of G that in't G itself, saying that though, I don't think I've conme across the term before either...*thinks*...

[bloodhound]

oh rite... maybe its equivalent to a "proper" subset. cheers .. ill have a go at doing this question.

[bloodhound]

oh rite... maybe its equivalent to a "proper" subset. cheers .. ill have a go at doing this question.

[matt grime]

what's your inkling? do you think it true or false?

[matt grime]

what's your inkling? do you think it true or false?

[Woxor]

At a guess bloodhound, I say a "proper" subgroup was just a subgroup of G that in't G itself, saying that though, I don't think I've conme across the term before either...*thinks*...

Yes, that's right. The only real conceptual difference between that and a proper subset is that the trivial subgroup is still a proper subgroup, whereas the null set isn't a proper subset.