Subgroup K Normal in Dn: Proof and Examples

[hsong9]

Homework Statement

Let Dn = {1,a,..a^n-1, b, ba,...ba^n-1} with |a|=n, |b|=2,
and aba = b.
show that every subgroup K of <a> is normal in Dn.

The Attempt at a Solution

First, we show <a> is normal in Dn.
<a> = {1,a,...a^n-1} has index 2 in Dn and so is normal
by Thm (If H is a subgroup of index 2 in G, then H is normal in G.)
Next, Since <a> is cyclic, K is also cyclic and abelian.
Let k in K, x in G and 1 in G. ( G = Dn)
k = 1k = (xx^-1)k = x(kx^-1) because K is abelian.
k in K => xkx^-1 in K for all x in G
=> K is a normal in G = Dn.

 

[latentcorpse]

are you sure the relation isn't meant to be $aba^{-1}=b$?

also i think it suffices to show that you have an index of 2 OR that for $n \in N, \forall g \in G, gng^{-1} \in N$ in order to imply $N \triangleleft G$.

 

[matt grime]

Let k in K, x in G and 1 in G. ( G = Dn)
k = 1k = (xx^-1)k = x(kx^-1) because K is abelian.

This doesn't follow. K is abelian, so elements of K commute with themselves; they do not commute with arbitrary elements of G. But then they do not need to - you're trying to show that K is normal, not central.