# Some abstract algebra help needed.

1. Nov 25, 2006

### 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!

2. Nov 25, 2006

### Hurkyl

Staff Emeritus
What have you tried? Where are you stuck?

3. Nov 25, 2006

### Playdo

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|.

Last edited: Nov 25, 2006
4. Nov 26, 2006

### 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?

Last edited: Nov 26, 2006
5. Nov 26, 2006

### HallsofIvy

Staff Emeritus
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 $\beta (y)$?

6. Dec 1, 2006

### Playdo

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.

7. Dec 1, 2006

### 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.)

Last edited: Dec 2, 2006