# abstract algebra

• Oct 4th 2008, 03:01 PM
dori1123
abstract algebra
Prove that if A is a subset of B, then <A> is a subgroup of <B>. Give an example where A is a subest of B with A =/= B but <A> = <B>.

• Oct 4th 2008, 03:59 PM
ThePerfectHacker
Quote:

Originally Posted by dori1123
Prove that if A is a subset of B, then <A> is a subgroup of <B>. Give an example where A is a subest of B with A =/= B but <A> = <B>.

$\displaystyle \left< A \right> = \bigcap_{A\subseteq H} H$ - the intersection of all subgroups of the group containing $\displaystyle A$.
$\displaystyle \left< B \right> = \bigcap_{B\subseteq H} H$ - the intersection of all subgroups of the group containing $\displaystyle B$.

Now since $\displaystyle A\subseteq B$ it follows,
$\displaystyle \left< A \right> = \bigcap_{A\subseteq H}H \subseteq \bigcap_{B\subseteq H} H = \left< B \right>$.

For the example, let $\displaystyle G = \mathbb{Z}_2$ then $\displaystyle \left< \{ [1] \} \right> = \left< G \right>$.
• Oct 4th 2008, 05:42 PM
dori1123
Quote:

Originally Posted by ThePerfectHacker
For the example, let $\displaystyle G = \mathbb{Z}_2$ then $\displaystyle \left< \{ [1] \} \right> = \left< G \right>$.

Is $\displaystyle \mathbb{Z}_2$ the same as Z/2Z?
• Oct 4th 2008, 06:00 PM
ThePerfectHacker
Quote:

Originally Posted by dori1123
Is $\displaystyle \mathbb{Z}_2$ the same as Z/2Z?

Yes