# Is the subgroup of an abelian group always abelian?

• Dec 4th 2009, 10:41 AM
MaryB
Is the subgroup of an abelian group always abelian?
That's my question. I think it is, but can you proof this?

And is it both ways: and if a group has a abelian subgroup, is the group itself always abelian?
• Dec 4th 2009, 10:45 AM
Bruno J.
Yes, no.

Think about it! If a group is abelian, then $\displaystyle xy=yx \forall x,y \in G$, and so this definitely holds for any subset $\displaystyle S \subset G$.

On the other hand, if you take any group $\displaystyle G$, then the trivial subgroup $\displaystyle \{1\}$ is abelian, but $\displaystyle G$ isn't necessarily.
• Dec 4th 2009, 10:58 AM
Jose27
For the second one it gets better: A group can have all its subgroups abelian without being abelian itself (Quaternion group)
• Dec 4th 2009, 11:20 AM
Bruno J.
Quote:

Originally Posted by Jose27
For the second one it gets better: A group can have all its (proper) subgroups abelian without being abelian itself (Quaternion group)

Yes, much better example!
• Dec 6th 2009, 03:14 AM
MaryB
Quote:

Originally Posted by Bruno J.
Yes, no.

Think about it! If a group is abelian, then $\displaystyle xy=yx \forall x,y \in G$, and so this definitely holds for any subset $\displaystyle S \subset G$.

On the other hand, if you take any group $\displaystyle G$, then the trivial subgroup $\displaystyle \{1\}$ is abelian, but $\displaystyle G$ isn't necessarily.

So, If a group is abelian, then $\displaystyle xy=yx \forall x,y \in G$, and so this definitely holds for any subset $\displaystyle S \subset G$ is a proof?
Thanks both!
• Dec 6th 2009, 10:06 AM
Bruno J.
Quote:

Originally Posted by MaryB
So, If a group is abelian, then $\displaystyle xy=yx \forall x,y \in G$, and so this definitely holds for any subset $\displaystyle S \subset G$ is a proof?
Thanks both!

What do you think? Does it convince you?
• Dec 6th 2009, 11:38 PM
Swlabr
Quote:

Originally Posted by MaryB
That's my question. I think it is, but can you proof this?

Also, if a group is abelian then so are all its homomorphic images (equivalently, quotients):

$\displaystyle (a\phi) (b \phi) = (ab)\phi = (ba)\phi = (b\phi) (a \phi)$