Share
Explore BrainMass

Cyclic group

Let X be a prime. Prove or disprove that is cyclic for each normal subgroup K.

See attached file for full problem description.

Attachments

Solution Preview

Proof:
First, we should know the definition of Dp.
Dp=<a,b> is generated by two elements a and b, where
a^2=b^p=1, ab=b^(-1)a.
So K=<b> is a normal subgroup of Dp. Because aba^(-1)=b^(-1) is in K.
Now I claim that K is the unique normal subgroup of Dp if p>=3 is a prime.
We know, |Dp|=2p, |K|=p, so [Dp:K]=2 and Dp = K union aK. So each ...

Solution Summary

This is a proof regarding a cyclic group.

$2.19