Basic Group Theory: Proof <A,B>n<C>=<AuBnC>

[tgt]

In a group G, is it true that <A,B>n<C>=<AuBnC> where A,B and C are sets in G?

Where <D> denotes the smallest subgroup in G containing the set D.

Proof
If g is in <A,B> and g is in <C> then g is capable of being generated by elements in A or B and also elements in C. So g is generated by elements in (AuB)nC. So g is in <(AuB)nC>.

if g is in <(AuB)nC> then g is in <AuB> and g is in <C> so g is in <AuB>n<C>

 

[CompuChip]

If g is in <A,B> and g is in <C> then g is capable of being generated by elements in A or B and also elements in C. So g is generated by elements in (AuB)nC. So g is in <(AuB)nC>.

I don't think that is true. For example, consider the additive groups generated by {3} and by {4} (i.e. 3Z and 4Z).
They both contain 12, yet the intersection of the generating sets is empty.

 

[tgt]

I don't think that is true. For example, consider the additive groups generated by {3} and by {4} (i.e. 3Z and 4Z).
They both contain 12, yet the intersection of the generating sets is empty.

nice one.

 

[tgt]

We can say that if A and B are subgroups of G then <A,B>=AuB, right?

 

[CompuChip]

Did you check with my example?
$$( \langle 3, 4 \rangle, + ) \neq (3\mathbb{Z} + 4\mathbb{Z}, +)$$
(in fact, the LHS is a group while the RHS is not).

 

[tgt]

Did you check with my example?
$$( \langle 3, 4 \rangle, + ) \neq (3\mathbb{Z} + 4\mathbb{Z}, +)$$
(in fact, the LHS is a group while the RHS is not).

I see.