Does every subgroup of an abelian group have to be abelian?

Question

My original problem is to show that E/L is an abelian extension over L and L/F is an abelian extension over F, given that E/F is an abelian extension over F and that L is a normal extension of F such that $F\subseteq L \subseteq E$.

So far I have proved that E is a normal extension of F, E is a normal extension of L, and L is a normal extension of F. I know that to prove abelian extension I must also prove that Gal(E/L) is an abelian group. I have shown that Gal(E/L) $\subseteq$ Gal (E/F). In my mind it makes sense that I cannot lose commutativity therefore my subgroup must be Abelian too. How do I show this in a proof? Is it enough to show two elements in the subgroup must also exist in the larger group and that they must be commutative in the larger group? I feel like I know what needs to be done, just not how to phrase it.

Answer 1

If $G$ is an abelian group and $H$ is a subgroup, suppose $x, y\in H$. Then in particular $x, y\in G$, so $xy=yx$. Since $x, y$ were arbitrary, $H$ is abelian.

Answer 2

Huh, funny, we just went over this today in my algebra class.

Yes, subgroups of abelian groups are indeed abelian, and your thought process has the right idea.

Showing this is pretty easy. Take an abelian group $G$ with subgroup $H$. Then we know that, for all $a,b\in H$, $ab=ba$ since it must also hold in $G$ (as $a,b \in G \ge H$ and $G$ is given to be abelian).

Answer 3

Just use proof by contradiction. Suppose $H$ is not abelian and thus contains two non-commuting members $x$ and $y$. Then $xy \neq yx$. But $x$ and $y$ are also in $G$, and thus $G$ is not abelian. Contradiction. Therefore $H$ is abelian.