# Cosets/normal subgroup

• Mar 9th 2008, 05:27 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, 05:36 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 $\displaystyle \forall n \in N$ and $\displaystyle \forall g \in G$, $\displaystyle gng^{-1} \in N$

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 $\displaystyle \forall g \in G$, $\displaystyle gN = Ng$

Quote:

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

Take $\displaystyle x,y \in H$

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

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

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

So yes, a subgroup of an abelian group is abelian. :)
• Mar 9th 2008, 06:49 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 $\displaystyle H$ of $\displaystyle G$ such that $\displaystyle ghg^{-1} \in H$ for all $\displaystyle g\in G$ and $\displaystyle h\in H$. Now suppose that $\displaystyle H$ is a subgroup that has this poperty. We will prove that $\displaystyle bH=Hb$ for any $\displaystyle b\in G$, by showing $\displaystyle bH\subseteq Hb \mbox{ and }Hb\subseteq bH$. Say $\displaystyle x\in bH$ then $\displaystyle x=bh_1$ but $\displaystyle bh_1b^{-1}\in H \implies bh_1b^{-1} = h_2 \implies bh_1 = h_2 b$ for some $\displaystyle h_2\in H$. And so $\displaystyle x=bh_1=h_2b\in Hb$. And the other way around is similar.