Is β a Homomorphism or Isomorphism in Abstract Algebra?

[harbong]

can anyone help me with my abstract algebra assignment?

Let a be an fixed element of some multiplicative group G. Define the map β: Z > G from the additive inter group Z to G by β(n)=a^n.
i. Prove that the map β is a homomorphism.
ii. Prove/Disprove that the map β is an isomorphism.

thanks!

 

[Hurkyl]

What have you tried? Where are you stuck?

 

[Playdo]

can anyone help me with my abstract algebra assignment?

Let a be an fixed element of some multiplicative group G. Define the map β: Z > G from the additive inter group Z to G by β(n)=a^n.
i. Prove that the map β is a homomorphism.
ii. Prove/Disprove that the map β is an isomorphism.

thanks!

WTH! Your terminology is a bit different than I am used to so I will state my assumptions. By fixed element of G you mean it is invariant under automorphism. Z meaning the integers under addition?

Ok, a^0 = 1 so your set is {1,a} at least. a will generate a multiplicative subgroup with $a^{\phi(|G|-1)}=1$ at most. Clearly beta is a homomorphism, but it cannot be an isomorphism because the kernel is the set $Ker(\beta)=\{0, t(|G|-1)| t \in Z\}$ at least. The order of the subgroup generated by a may divide |G|.

 

[matt grime]

@playdo

Of course it *can* be an isomorphism. It is just not *necesarily* an isomorphism.

I don't know where you get \phi(|G|-1) from. What you wrote there implies that |G| and \phi(|G|-1) are not coprime, which seems very unlikely (take |G|=p your favourite odd prime). The orders of elements divide |G|, and the order of a could very well be |G|, as you even half imply yourself, so how on Earth \phi(|G|-1) comes into it is not clear at all.

Fixed, in this context, just means 'take some choice of a', it does not mean 'invariant under automorphism - what automorphism?

 

[HallsofIvy]

Harbong, what is the definition of "homomorphism" and "isomorphism"? Show that the specific requirements in the definition of homomorphism are satisfied but not those of isomorphism. In particular, for a homomorphism, we must have $\beta (x+ y)= \beta (x)\beta (y)$. If x and y are arbitrary integers, what is $\beta (x+ y)$? What are $\beta (x)$ and $\beta (y)$?

 

[Playdo]

Fixed, in this context, just means 'take some choice of a', it does not mean 'invariant under automorphism - what automorphism?

That's why I asked Dummit and Foote ( an often used textbook in graduate schools) defines fixed element to be an element of a group G that is left fixed by some automorphism of G.

 

[Playdo]

Your right I was thinking modular and got the formula wrong. This is not strictly modular though. I guess to satisfy I will just do the damn thing right.

A homomorphism (M) between two groups Z (integers under addition) and G (multiplicative group) has the following properties.

i) M(x+y) = M(x)M(y)

ii) M(0) = 1

iii) M(~x) = ~M(x) - ~ here is meant to imply inverses.

Now clearly

ii) M(0) = a^0 = 1 for all a >< 0 in G

i) M(x+y) = a^(x+y) = a^x*a^y= M(x)M(y)

iii) M(-x)M(x) =a^(-x)a^(x) = a^(-x+x)=a^0=1

So it is a homomorphism for all a not equal to zero. Whic is fortunate for us because zero just does not fit into the idea of a multiplicative group except for the singleton group {0,*} whic is structurally identical to the singelton group {1,*}. However I don't want to assume that multiplication in G is standard in any sense.

The isomorphism part we have to show that the map M is one to one. But that is just the same as saying the kernel of M is precisely 0. However the multiplicative group that a is coming from is not specified so there are a few alternatives. Before we proceed remember that for there to be an isomorphism |Z| must equal |G|.

By definition the kernel of M is {x in Z|M(x) = a^x = 1}. This is the key to understanding the possible maps. We need to look at the possible multiplicative groups.

1) |G| = |Z| -> M(Z,a) = {...,a^-3,a^-2,a^-1,1,a,a^2,a^3,...}. To find the kernel invert M(x,a) = a^x = 1. Clearly x = log[a](1) = 0 is the only solution. So this option leaves us with <G,*> isomorphic to <Z,+> and to some multiplicative subgroup of <Q+,*>.

2) |G|= K a natural number. So now we could have any reduced residue system as G, in which case a coprime to K is required for a to generate a subgroup of G otherwise it generates a subset of zero divisors that may or may not have any meaningful structure. But that won't happen, every a in the reduced residue system mod |G| will be coprime with |G|. These maps are not isomorphisms of Z into G and are covered under the heading of the fundamental theorem of group homomorphisms. The book by Fraleigh handles it as well as any. I had thought it might require the Sylow Theorems, but that just helps us understand the number and size of subgroups when |G| = mp^n where m and p are coprime.

The basic overarching fact is that you cannot have an isomorphism between sets that are of different sizes. I think the only property not dealt with here is whether G is abelian. Certainly Z is abelian, is it possible that G is not abelian?

That is to say

M(x+y) = M(x)M(y) >< M(y)M(x) = M(y+x)

It has been a while since I took abstract algebra but this statement seems to preclude G being non-abelian since it implies that x+y >< y+x which in this case is false. In fact it would seem you can never have an isomorphism between abelian and non-abelian groups. Could we have a homorphism? Clearly no. So I think that should about do it, we don't need to worry about the non-abelian groups D[k] or S[k] (dihedral and symmetric groups of order k.)