1. Abstract Alg.-Abelian groups presentation

Let Cn be a cyclic group of order n.
A. How many sub-groups of order 4 there are in C2xC4... explain.
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?
D. What is the Automorphism group of an infinite cyclic group?

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

2. 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 $C_2:=\,,\,\,C_4:=$, then it's easy to check that the following are sbgps. of order 4 of $C_2\times C_4$ :

$\{1\}\times C_4=\{(1,b),(1,b^2),(1,b^3),(1,1)\}\,,\,\,<(a,b)>= \{(a,b),(1,b^2),(a,b^3),(1,1)\}$ , $\times=\{(1,1),(1,b^2),(a,1),(a,b^2)\}$

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 $H=$ , then any $\phi \in Aut(H)$ is completely and uniquely determined by $\phi(h)$ , and since this must be an AUTOMORPHISM it follows that $\phi(h)$ has to be a generator of the group, so $\phi(h)=h^r$ , with $(r,8)=1\Longleftrightarrow\,r=1,3,5,7\Longrightarr ow |Aut(H)|=4$.

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 $\mathbb{Z}$ 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.

.

3. 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?