NEED A PERFECT PAPER? PLACE YOUR FIRST ORDER AND SAVE 15% USING COUPON:

4.8/5

MATH 421 Miami University Abelian Group and Cyclic Group Questions

MATH 421 Miami University Abelian Group and Cyclic Group Questions.

Question Description

HW 2 is due Sunday September 6, by 11:59pm. Please upload your solutions on canvas under
Assignments by the due time.

1. Let G be a group and H and F two different proper subgroups of G.

(a) Prove that H ∩ F is also a subgroup of G.
(b) Prove that if H and F do not contain each other then H ∪ F is not a subgroup of G.
(hint: by our assumption there is an element a that lies in H but not in F and an element that
lies in F but not in H. Suppose H ∪ F were a subgroup of G, derive a contradiction through a, b.)

2. (a) Let G be an abelian group and H = {x ∈ G : |x| is odd}. Prove that H is a subgroup of G.
(b) Let G be a group and a an element of G. Prove that C(a) ⊆ C(an) for every n ∈ Z+, where

C(x) denotes the centralizer of x in G.
3. Let m,n be elements of the additive group Z. Determine a generator (with justification) of

⟨m⟩ ∩ ⟨n⟩.

4. Let G be a group. Let p, q be two different prime numbers. Suppose a, b ∈ G are elements in
G of order p and q respectively. Prove that ⟨a⟩ ∩ ⟨b⟩ = {e}. Do not use any knowledge beyond
Chapter 4 on cyclic subgroups.

5. Let G be a cyclic group. Prove that G cannot be expressed as the union of its proper subgroups.
6. (Graduates only) Let G be a group and a,b ∈ G. Suppose that |a| = 12 and |b| = 22 and

⟨a⟩∩⟨b⟩≠ {e}. Provethata =b .

MATH 421 Miami University Abelian Group and Cyclic Group Questions

Solution:

Why US?

100% Confidentiality

Information about customers is confidential and never disclosed to third parties.

Timely Delivery

No missed deadlines – 97% of assignments are completed in time.

Original Writing

We complete all papers from scratch. You can get a plagiarism report.

Money Back

If you are convinced that our writer has not followed your requirements, feel free to ask for a refund.

WhatsApp us for help!
%d bloggers like this: