# Cosets/normal subgroup

• Mar 9th 2008, 06:27 PM
hercules
Cosets/normal subgroup
Can someone please explain to me normal subgroups.

if you have H a subgroup of an abelian group G and b is an element of G, then the right coset to which b belongs is same as the left coset to which b belongs. (proving Hb=bH).

also, is a subgroup of an abelian group, necessarily abelian?

Thank you
• Mar 9th 2008, 06:36 PM
WWTL@WHL
Hello :)

Quote:

Originally Posted by hercules
Can someone please explain to me normal subgroups.

Take a group, G. then N is called a normal subgroup of G iff $\forall n \in N$ and $\forall g \in G$, $gng^{-1} \in N$

You can also say N is a normal subgroup of G if the sets of left and right cosets of N in G coincide, or $\forall g \in G$, $gN = Ng$

Quote:

also, is a subgroup of an abelian group, necessarily abelian?
Say H is a subgroup of an abelian group, G.

Take $x,y \in H$

Then since H is a subgroup, $x,y \in G$

Since G is abelian, you can say $x*y=y*x$

So $x*y = y*x \in H$ by closure

So yes, a subgroup of an abelian group is abelian. :)
• Mar 9th 2008, 07:49 PM
ThePerfectHacker
Quote:

Originally Posted by hercules
if you have H a subgroup of an abelian group G and b is an element of G, then the right coset to which b belongs is same as the left coset to which b belongs. (proving Hb=bH).

There are different ways to define normality, the way I perfer (and most algebra books use) is a subgroup $H$ of $G$ such that $ghg^{-1} \in H$ for all $g\in G$ and $h\in H$. Now suppose that $H$ is a subgroup that has this poperty. We will prove that $bH=Hb$ for any $b\in G$, by showing $bH\subseteq Hb \mbox{ and }Hb\subseteq bH$. Say $x\in bH$ then $x=bh_1$ but $bh_1b^{-1}\in H \implies bh_1b^{-1} = h_2 \implies bh_1 = h_2 b$ for some $h_2\in H$. And so $x=bh_1=h_2b\in Hb$. And the other way around is similar.