Thread: Abelian and normal Subgroups

I came across an exercise on Abelian groups which said that,
"In an abelian group G, every subgroup is normal". I had no issues proving this result. But my question is this.

Re: Every subgroup of an abelian group G is normal.
# 1. Does this mean that the group G, an improper subgroup of itself, is also normal?
# 2. Does this mean that commutativity implies normality of groups?
Is this similar to saying that every cyclic group is abelian?
# 3. If 2 is not true, are there any specific conditions under which 2 is true or false?

Originally Posted by MAX09

I came across an exercise on Abelian groups which said that,
"In an abelian group G, every subgroup is normal". I had no issues proving this result. But my question is this.

Re: Every subgroup of an abelian group G is normal.
# 1. Does this mean that the group G, an improper subgroup of itself, is also normal?

yes

# 2. Does this mean that commutativity implies normality of groups?

isn't this just rephrasing the same question? if a group is commutative, it is Abelian, and if it's Abelian, all its subgroups are normal.

Is this similar to saying that every cyclic group is abelian?

not really. here we are talking about the group itself. but the theorem in question goes beyond that. not only do we know something about the group itself, but every subgroup of that group. i suppose the universal quantifier makes it similar. the "every" part.

1) Groups aren't normal, subgroups are. G is a normal subgroup of itself for any group, abelian or not. That's why a simple group is defined as a non-trivial group with no normal subgroups other than the trivial group and the group itself (see Wikipedia for example)
2) Any subgroup of an abelian (i.e. commutative) group is a normal subgroup. As before groups aren't normal - the same group can easily be a normal subgroup in some groups but non-normal in others. All cyclic groups are abelian but not all abelian groups are cyclic.

Thanks, it helped