Is the subgroup of an abelian group always abelian?

**MaryB** · December 4th 2009, 10:41 AM

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?

**Bruno J.** · December 4th 2009, 10:45 AM

Yes, no.

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

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

**Jose27** · December 4th 2009, 10:58 AM

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

**Bruno J.** · December 4th 2009, 11:20 AM

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!

**MaryB** · December 6th 2009, 03:14 AM

Originally Posted by Bruno J.

Yes, no.

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

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

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

**Bruno J.** · December 6th 2009, 10:06 AM

Originally Posted by MaryB

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

What do you think? Does it convince you?

**Swlabr** · December 6th 2009, 11:38 PM

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):

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

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