Why Choose Us?
0% AI Guarantee
Human-written only.
24/7 Support
Anytime, anywhere.
Plagiarism Free
100% Original.
Expert Tutors
Masters & PhDs.
100% Confidential
Your privacy matters.
On-Time Delivery
Never miss a deadline.
Let G be a nonabelian group and Z(G) be its center
Let G be a nonabelian group and Z(G) be its center. Show that the factor group G/Z(G) is not a cyclic group.
We know if G is abelian, Z(G)=G. But now if it is not abelian, can we simply say because G is not cyclic, then any factor group will not be cyclic either? or is there more to it?
Expert Solution
Proof:
If the factor group G/Z(G) is cyclic, then G/Z(G)=<gZ(G)> for some g in G. So for each element in G, it has the form (g^k)a, where a is in Z(G).
Now we consider any two elements (g^m)a and (g^n)b in G, where a,b are in Z(G). So a and b are commutative to any elements in G. Then we have
(g^m)a * (g^n)b
=(g^m)(a*g^n)b
=(g^m)(g^n)ab
=(g^n)(g^m)ba
=(g^n)b * (g^m)a
So G is abelian. But from the condition, G is not abelian. We get a contradiction.
Therefore, G/Z(G) can not be cyclic.
Archived Solution
You have full access to this solution. To save a copy with all formatting and attachments, use the button below.
For ready-to-submit work, please order a fresh solution below.
