Few problems related to abstract algebra

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Hi everyone;
There are some questions which are frizzling my mind, if anybody could help then please reply to these ques which are as follows.
Q1) Prove that homomorphic image of cyclic group is itself cyclic?

Q2) Prove that any group 'G' can be embedded in a group of bijective mapping of a certain set? ( Here bijective mapping is that which is one-to-one as well as onto i.e. injective and surjective both).

Q3) Prove that the number of elements in a conjugacy class Ca of an element 'a' in a group 'G' is equal to the index of its normalizer?

If anyone has an idea about any of the above proofs then please let me know, I hope u guys will give me this favor.

Bye
 
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Q1)
Let G,G' be groups and let \phi : G \rightarrow G' be a homomorphism. Let g be a generator for G.

First of all you want to show that \phi (G) is a subgroup of G'. How do you show a subset of a group is a subgroup?

Next you want to find a generator for \phi (G).

Q2)
Let x \in G. Consider the function \lambda_x : G \rightarrow G defined by g \mapsto xg for all g \in G. (\lambda_x can be thought of as performing left multiplication by x)

Can you show \lambda_x is a bijection on G?

Consider the set of all such mappings, ie:

H = \{ \lambda_x : x \in G \} [/itex]<br /> <br /> Is H a group?
 
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1) Just do it; it follows instantly from the definitions

2) What is the only set you know any abstract group may act on?

3) This is straight forward from the orbit-stabilizer theorem, which I shall cite just in case you've not had it labelled by that name: if G is a finite group acting on a set X then |G| = |Orbit(x)||Stab(x)| for any x in X. Or you can do it by counting cosets of the normalizer, which is essentially reproving this statement for a specific case.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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