Thread: Abstract Alg.-Abelian groups presentation

Originally Posted by WannaBe

Let Cn be a cyclic group of order n.
A. How many sub-groups of order 4 there are in C2xC4... explain.


Put , then it's easy to check that the following are sbgps. of order 4 of :

,

Try now to mimick the above with the group below.
B. How many sub-groups of order p there are in CpxCpxC(p^2) when p is a prime? explain.
C. Prove that if H is cyclic of order 8 then Aut(H) is a non-cyclic group. WHAT is its order?


If , then any is completely and uniquely determined by , and since this must be an AUTOMORPHISM it follows that has to be a generator of the group, so , with .

Now check that this can't be a cyclic group, say by finding two different elements (i.e., automorphisms) of order 2...


D. What is the Automorphism group of an infinite cyclic group?

Just as above: the image of a generator MUST be a generator, so....what are the generators of and what're the possibilities of homomorphisms between them...


About A-> it's obvious (because it has index 2) that a subgroup of order 4 is normal...

Oh, dear: much easier! EVERY sbgp. of an abelian group is trivially normal...

Tonio


But I can't figure out how many subgroups of this form there are...
About the other parts-I've no idea...

I'll be delighted to get guidance about all the parts in this question.

Tnx in advance...

.

Hey tonio, 10x for you answer.... But:
About 1: How did you get to this result? How can we prove that there are no more order 4 sbgps of C2xC4...? Is there a non-brute-force way to check it out?

About 2: an order p sbgrp is cyclic of course...Hence we need to find all the (a,b,c) in CpxCpxCp^2 such as
lcm(o(a),o(b),o(c)) = p... p is a prime so one of the elements must have order p and the other two order 1 or p... The options are:
p, p, p & p,1,1& 1,p,1& 1,1,p& p,p,1 etc... and these are the only options...
AM I right?

About 3: Completely understandable...TNX

About 4: Let H be an infinite cyclic group. Hence H is isomorphic to Z. Let H=<h>... Then we can check the Aut(Z)... Each aut. of Z is determined by the image of the generators of Z... The only generators of Z are 1 and -1... SO there are these possibilities: 1->1, -1->-1 = Identitiy aut. or 1-> -1 , -1->1 and that's it... SO there are 2 automorphisms only...Am I right?

Your verification& help is needed as you can see... espacially in parts 1&2...

TNX in advance!